First solution. Note that the difference between the roots of a monic quadratic trinomial \( x^{2} + bx + c \) is equal to the square root of its discriminant, i.e., \( \sqrt{b^{2}-4c} \).

Let the two given trinomials be \( f(x) = x^{2} + b_{1}x + c_{1} \) and \( g(x) = x^{2} + b_{2}x + c_{2} \). According to the condition, they have a common discriminant \( D = b_{1}^{2}-4c_{1} = b_{2}^{2}-4c_{2} \). Instead of the sum of the trinomials, it is convenient to consider their half-sum, which is also a monic quadratic trinomial. The square of the difference of its roots (i.e., the discriminant) is equal to

\[ \left(\frac{b_{1} + b_{2}}{2}\right)^{2} - 2\left(c_{1} + c_{2}\right) = \frac{b_{1}^{2} + b_{2}^{2}}{2} - 2\left(c_{1} + c_{2}\right) - \left(\frac{b_{1}-b_{2}}{2}\right)^{2}. \]

Thus, it is not greater than \( \frac{b_{1}^{2} + b_{2}^{2}}{2} - 2\left(c_{1} + c_{2}\right) = \frac{D}{2} + \frac{D}{2} = D \). Hence, it follows that the difference between the roots of the half-sum is not greater than \( \sqrt{D} \), which is the difference between the roots of each of the given trinomials.

Remark. In the above estimate, one could use the inequality of means \( \sqrt{\frac{b_{1}^{2} + b_{2}^{2}}{2}} \geqslant \frac{b_{1} + b_{2}}{2} \).

Second solution. Note that any monic quadratic trinomial with two roots has the form \( (x - p)^{2} - q^{2} \) for \( q \geqslant 0 \). In this case, the difference between its roots is \( 2q \), and its minimum value is \( -q^{2} \).

Now the condition means that the two given trinomials have equal minimum values \( -q^{2} \). The minimum value of their half-sum is obviously not less than \( -q^{2} \) (it is the half-sum of some values of the original trinomials), i.e., it is equal to \( -r^{2} \) for \( 0 \leqslant r \leqslant q \). Therefore, the difference between the roots of the half-sum, i.e., \( 2r \), does not exceed \( 2q \).

Remark. Practically the same reasoning shows that the statement of the problem remains true if, instead of two monic trinomials, we consider two arbitrary quadratic trinomials with positive leading coefficients.

First solution. Number the positions in the string from left to right with numbers from 1 to 250. Let there be \( x \) letters A in the original string at odd positions (i.e., positions with odd numbers). We will show that in the resulting strings, this number does not change.

Indeed, suppose for some operation a segment is chosen in which there are \( y \) letters A and B, with \( t \) of these letters A at odd positions. Then there are \( y - t \) letters A at even positions in the segment, and therefore \( y-(y - t) = t \) letters B. After the operation, it is from these \( t \) letters B that letters A will arise, standing at odd positions in the segment—thus, the number of such letters A does not change.

Thus, in any resulting string, there will be exactly \( x \) letters A at odd positions. However, if the string is reversed, then at odd positions there should be exactly those letters that were previously at even positions, where there were exactly \( 125-x \) letters A. Since \( 125-x \neq x \), the required result is impossible.

Note. The solution above presents an invariant of the process (i.e., a quantity that remains constant)—the number of letters A at odd positions. There are other similar invariants that allow solving the problem. For example, one can show that the sum of the positions of the letters A is such an invariant.

Second solution. We present another invariant. In the string, there are a total of \( 125^{2} \) pairs consisting of a letter A and a letter B. We call such a pair left if A is to the left of B, and right otherwise. We will show that during the operation, the number of left pairs does not change. From this, it will follow that the required result is impossible, because when the string is reversed, all pairs change type, and therefore the number of left pairs changes parity.

Consider one operation with a segment of length \( 2 y \). In this operation, pairs of letters not in the segment retain their type. Furthermore, for each letter outside the segment, there were exactly \( y \) pairs containing it and a letter from the segment; the same number of such pairs remain, and all these pairs were and became of the same type.

Thus, it remains to track the pairs of letters in the segment itself. But each pair changed its type twice: when the segment was reversed and when all letters were replaced by others. Therefore, the number of left pairs in the segment also did not change.

Let \( ABC \) be the given triangle, \( O \) be the center of its circumcircle, and \( D, E, F \) be the midpoints of its sides \( BC, CA, AB \) respectively, such that \( DEF \) is similar to \( ABC \) with a ratio of \( 1/2 \) and \( OD \perp BC, OE \perp CA, OF \perp AB \).

Let the point \( D \) move to \( D' \) when rotated around \( O \) clockwise by an angle of \( 120^{\circ} \). Under such a rotation, the line \( BC \) moves to a perpendicular to \( OD' \) passing through \( D' \), let this perpendicular intersect \( BC \) at point \( K \) (see image 3). We see that the right triangles \( ODK \) and \( OD'K \) are equal (symmetric with respect to \( OK \)), and therefore \( \angle KOD = \angle DOD'/2 = 60^{\circ} \), which means in the right triangle \( KOD \) it holds that \( OK = 2OD \). In other words, \( K \) is obtained from \( D \) as a result of a rotational homothety: a rotation with center \( O \) clockwise by an angle of \( 60^{\circ} \) followed by a homothety with center \( O \) and ratio 2. A similar result is obtained for the other points \( L, M \) of intersection of the lines of the same color. Thus, the triangle \( KLM \) is obtained from \( DEF \) by a rotational homothety with center \( O \) and ratio 2. Then \( KLM \) is similar to \( DEF \) with a ratio of 2, hence equal to \( ABC \).

On the 1st step, each of the 100 people wrote down one of the numbers from the set \( A_{1} = \{0,1,2, \ldots, 100\} \).

On the 2nd step, one of the numbers from the set \( A_{2} = \left\{\frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \ldots, \frac{199}{2}\right\} \).

On the 100th step, one of the numbers from the set \( A_{100} = \left\{\frac{S}{100}, \frac{S - 1}{100}, \frac{S - 2}{100}, \ldots, \frac{S - 100}{100}\right\} \), where \( S = \frac{100 \cdot 101}{2} \) is the sum of all numbers (and the subtracted number is the number on the remaining card at the end).

We see that \( A_{1} \cup A_{2} = \left\{0, \frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \ldots, \frac{199}{2}, \frac{200}{2}\right\} \), so \( \left|A_{1} \cup A_{2}\right| = 201 \). Further, \( \left|A_{100}\right| = 101 \), but the numbers \( 50-\frac{1}{2}, 50, 50 + \frac{1}{2} \) belong to \( A_{2} \cap A_{100} \), hence \( \left|A_{1} \cup A_{2} \cup A_{100}\right| \leqslant 201 + 101 - 3 = 299 \).

Thus, we have shown that the 300 numbers written on the 1st, 2nd, and 100th steps can take no more than 299 different values. Therefore, some two of them are equal.

For \( n = 20! \), we have \( \frac{n(n + 1)(n + 2) \ldots(n + 20)}{n^{2}} = \frac{(n + 1)(n + 2) \ldots(n + 20)}{20!} = C_{n + 20}^{20} \), which is an integer.

Now, let \( n > 20! \) and let \( P = n(n + 1)(n + 2) \ldots(n + 20) \) be divisible by \( k^{2} \), where \( k = n + i, i \in\{0,1,2, \ldots, 20\} \). We have \( P / k = (k - i)(k - i + 1) \ldots(k - 1)(k + 1)(k + 2) \ldots(k + j) \), where \( j = 20-i \). Notice that the number \( P / k \equiv(-1)^{i} i!j!(\bmod k) \) must be divisible by \( k \). But \( 0 < i!j!\leqslant i!\cdot(i + 1)(i + 2) \ldots(i + j) = 20! < n \leqslant k \), hence \( i!j! \) is not divisible by \( k \). Contradiction.

Let \( a = \sin x + \operatorname{tg} x \) and \( b = \cos x + \operatorname{ctg} x \). Introduce the notations: \( u = \sin x + \cos x \) and \( v = \sin x \cdot \cos x \). According to the condition, the numbers \( c = a + b = u + \frac{1}{v} \) and \( d = a \cdot b = v + u + 1 \) are rational. Hence, \( k = d - c = v + 1 - \frac{1}{v} \). Therefore, \( t = \sin 2x = 2v \) is a root of the quadratic equation \( t^{2} + 2t - (4 + 2kt) = 0 \) with rational coefficients, which implies the required result.

From the problem statement, we immediately obtain that \( \angle XYZ = 90^{\circ} = \angle XTZ \). Let \( Q \) be the intersection point of the lines \( XY \) and \( ZT \), and \( R \) be the intersection point of the lines \( ZY \) and \( XT \) (see image 7). Without loss of generality, we can assume that point \( Z \) lies on the segments \( RY \) and \( QT \). Since point \( R \) lies in both the plane \( ABD \) and the plane \( BCD \), it lies on the line \( BD \). Similarly, point \( Q \) lies on the line \( AC \).

Image 7

Image 8

Note that \( RY \) and \( QT \) are the altitudes of triangle \( XQR \). Then \( Z \) is the intersection point of the altitudes of this triangle, and therefore \( XZ \perp QR \). Let \( M \) be the midpoint of the segment \( QR \). Since \( \angle QYR = 90^{\circ} \), it follows that \( YM = MR = RQ \) by the property of the median in a right triangle. Thus, \( \angle MYR = \angle YRQ = 90^{\circ} - \angle XQR = \angle ZXQ \). Consequently, the line \( YM \) is tangent to the circle \( \omega \). Similarly, the line \( TM \) is also tangent to the circle \( \omega \), so the points \( M \) and \( P \) coincide.

Consider two parallel planes \( \beta \) and \( \gamma \), one of which contains the segment \( AC \), and the other contains the segment \( BD \). Note that the midpoints of all segments connecting a point from plane \( \beta \) and a point from plane \( \gamma \) lie in a single plane parallel to \( \beta \) and \( \gamma \). Indeed, if we introduce Cartesian coordinates such that one of the planes is given by the equation \( z = 0 \), and the other by the equation \( z = h \) (where \( h \) is the distance between the planes \( \beta \) and \( \gamma \)), then the midpoints of all considered segments lie in the plane \( z = h / 2 \). Applying this observation to the segments \( AB, BC, CD, DA, QR \), we find that their midpoints lie in the same plane, as required.

First solution. Suppose this is possible.

Lemma. Let the number \( n \) be blue; then \( n^{2} \) is red.

Proof. Since \( n \) is blue, the numbers \( n - 1 \) and \( n + 1 \) are red, otherwise two blue numbers differ by 1. Therefore, the number \( n^{2}-1 = (n - 1)(n + 1) \) is blue. Hence, \( n^{2} \) is red.

Obviously, there exists a blue number \( k > 1001 \); by the lemma, the number \( k^{2} \) is red. Consider two cases.

Let the number \( k^{3} \) be blue. Then by the lemma, the number \( k^{6} = \left(k^{3}\right)^{2} \) is red. Since \( k^{2} \cdot k^{4} = k^{6} \), and the numbers \( k^{2} \) and \( k^{6} \) are red, the number \( k^{4} \) must be blue. By the lemma, the number \( k^{8} = \left(k^{4}\right)^{2} \) is red, but it is the product of the red numbers \( k^{2} \) and \( k^{6} \). This cannot be.

Now let the number \( k^{3} \) be red. Then the number \( k^{5} = k^{2} \cdot k^{3} \) is blue, and by the lemma, the number \( k^{10} = \left(k^{5}\right)^{2} \) is red. Since \( k^{10} = k^{2} \cdot k^{8} = k^{3} \cdot k^{7} \) and the numbers \( k^{2} \) and \( k^{3} \) are red, the numbers \( k^{7} \) and \( k^{8} \) must be blue, and then, by the lemma, the numbers \( k^{14} = \left(k^{7}\right)^{2} \) and \( k^{16} = \left(k^{8}\right)^{2} \) are red. But then the red number \( k^{16} \) equals the product of the red numbers \( k^{2} \) and \( k^{14} \). Contradiction.

Second solution. Again, assume the contrary. Let's start with the following remark. Let \( a \) and \( b \) be two distinct red numbers; then the number \( a b + 1 \) is also red. Indeed, by the condition, the number \( a b \) is blue, and then \( a b + 1 \) is red.

The colors of the numbers cannot strictly alternate—otherwise, all numbers of one parity would be red, and then there would be two red numbers with a red product. Thus, there are two numbers of the same color differing by 1—let this be \( a \) and \( a + 1 \). By the condition, their common color is red.

From the above remark, we first obtain that the number \( b = a^{2} + a + 1 = a(a + 1) + 1 \) is red, and then that the number \( c = a^{3} + a^{2} + a + 1 = a b + 1 \) is also red. Hence, by the condition, the number \( d = a^{4} + 2 a^{3} + 2 a^{2} + 2 a + 1 = (a + 1) c \) is blue.

On the other hand, from the same remark, the number \( p = (a + 1) b + 1 = a^{3} + 2 a^{2} + 2 a + 2 \) is red. Hence, by the condition, the number \( a p = a^{4} + 2 a^{3} + 2 a^{2} + 2 a = d - 1 \) is also blue. Thus, we found two consecutive blue numbers \( d \) and \( d - 1 \), which is impossible.

Let \( \Gamma \) be the circle with diameter \( I_{B} I_{C} \). Since \( C I_{C} \perp C I_{B} \) and \( B I_{C} \perp B I_{B} \), points \( B \) and \( C \) lie on \( \Gamma \) (see Fig. 1).

Let \( I \) be the incenter of \( ABC \). If point \( X \) lies inside the angle \( BIC \), then angles \( XBI_{C} \) and \( XCI_{B} \) are obtuse, so \( X I_{B} > X C \) and \( X I_{C} > X B \). Multiplying these inequalities, we obtain the required result.

Otherwise, points \( X \) and \( A \) lie in the same half-plane with respect to line \( BC \). Extend rays \( CX \) and \( I_{B}X \) to intersect \( \Gamma \) at points \( C_{1} \) and \( Y \) respectively. Since quadrilateral \( AI_{C}BI \) is inscribed in a circle with diameter \( II_{C} \), we have \( \angle XC_{1}B = \angle II_{C}B = \angle IAB = \frac{1}{2} \angle CAB < \frac{1}{2} \angle CXB = \frac{1}{2}(\angle XC_{1}B + \angle XBC_{1}) \), from which \( \angle XC_{1}B < \angle XBC_{1} \), hence \( XC_{1} > XB \). Moreover, since the length of a chord of a circle does not exceed the length of the diameter, \( I_{B}X + XI_{C} \geq I_{B}I_{C} \geq I_{B}Y = I_{B}X + XY \), from which \( XI_{C} > XY \). Therefore, \( X I_{B} \cdot X I_{C} \geq X I_{B} \cdot XY = XC \cdot XC_{1} > XC \cdot XB \).

It is easy to see that if \( b = 1 \), then any polynomial with coefficients from 0 to \( b - 1 \) is the zero polynomial.

Let \( b > 1 \). Represent \( a - b \) in base \( b \) notation: \( a - b = c_{n} b^{n} + \ldots + c_{1} b + c_{0} \), where \( c_{i} \in \{0,1,2, \ldots, b - 1\} \). Since \( a - b \geqslant b \), in this representation \( n \geqslant 1 \).

We will show that \( P(x) = c_{n} x^{n} + \ldots + c_{1} x + c_{0} \) satisfies the condition. Indeed, for any polynomial \( f \) with integer coefficients, \( f(a) - f(b) \) is divisible by \( a - b \). Therefore, \( P(a) - P(b) \) is divisible by \( a - b = P(b) \). But then \( P(a) = (P(a) - P(b)) + P(b) \) is also divisible by \( P(b) \).

Solution. We will represent the tournament as an \( n \times n \) table, in which both the columns and the rows are numbered from 1 to \( n \). The columns correspond to the girls, and the rows to the boys. Then each game is represented by a cell whose coordinates correspond to the numbers of the girl and the boy playing in that game. First, place a token in the cell \( (1,1) \). After a girl’s victory, the token moves upward, and in the case of a boy’s victory, it moves to the right. If the token reaches the edge of the table, then from the last row when moving upward it moves to the first row, and from the last column when moving to the right it moves to the first column. Then the condition of the problem is equivalent to the token visiting all the cells of the table, being in each of them exactly once.

Color the cells of the table in \( n \) colors along the diagonals going right-down: the first diagonal in the first color, the second in the second, …, the \( n \)-th diagonal in the \( n \)-th color, and the following diagonals again in colors from the first to the \( n-1 \)-th. Note that after each game, the color number of the cell in which the token is located increases by 1 modulo \( n \). Since a total of \( n^{2} \) games were played in the tournament, which is divisible by \( n \), the token in the end is in a cell of the \( n \)-th color, that is, on the main diagonal (hereafter, when we say “diagonal,” we will mean precisely this diagonal). Let the final cell in the token’s route be located in the column with number \( m \); then it is required to prove that the number \( m \) is odd.

From the top cell of the diagonal, the token could not go upward, since it had already been in the cell \( (1,1) \). Hence, if this cell is not final, then from it the token went to the right. Then from the next cell of the diagonal it also made a move to the right, and so on, up to the cell located in the column with number \( m-1 \). Similarly, from the cells of the diagonal located in the columns with numbers from \( m+1 \) to \( n \), the token moved upward (see fig. 4). Let the first cell of the diagonal that the token entered be in the column with number \( k \).

Consider the path of the token from the starting cell to this cell. All paths from cells of the first color to the next cell of color \( n \) must be the same as the path under consideration; namely, each such path is obtained from another by shifting by the vector \( (1,-1) \). Indeed, if the token from the cell \( (a-1, b) \) made a move upward, and from the cell \( (a, b-1) \) to the right, then it would not arrive at the cell \( (a, b) \); and if from these cells it made moves to the right and upward respectively, then it would arrive at one and the same cell twice. Therefore, from each such pair of cells the token made the same moves.

Without loss of generality, we will assume that \( k<m \). The cells of the diagonal that are to the left of the final cell will be called left, and those that are to the right — right. We number the left cells by the numbers from 1 to \( m-1 \), and the right cells — from 1 to \( n-m \) (in both cases we number them moving right-down). Let us see in what order the token visited these cells. From the left cells it shifted \( k \) cells to the right (since from them it made a move to the right to reach a cell of the first color), and from the right cells — \( k-1 \) cell(s) to the right. Hence, for left cells only the remainder of the number upon division by \( k \) is important, and for right cells — the remainder upon division by \( k-1 \). Moreover, if the number of right cells is less than \( k \), one may increase \( n \) by \( 2(k-1) \), adding \( 2(k-1) \) right cells; this will not affect the subsequent argument. For convenience, we replace all cell numbers by the corresponding remainders, and for right cells, instead of the remainder 0, we use the number \( k-1 \).

Let the number \( m \) give remainder \( d \) when divided by \( k \). Then the first transition from left cells to right cells was from the number 0 to the number \( k-d \), and at this moment all cells with zero in the left part were visited. Only the numbers from 1 to \( k-1 \) remain on the diagonal. Next, the chain of transitions between right and left cells looks like \( k-d \rightarrow \cdots \rightarrow d \). In this chain each number from 1 to \( k-1 \) occurs twice; it begins on right cells and ends on left cells. Transitions from right cells to left cells we call transitions of the first type, and from left to right — of the second type. Then the chain contains \( k-1 \) transitions of the first type and \( k-2 \) of the second type, and they alternate.

We prove that every two numbers in the chain, symmetric with respect to its center, sum to \( k \). This is true for the extreme numbers. Every two symmetric transitions have the same type; therefore, in them the same number is added modulo \( k-1 \) (for transitions of the first type) or modulo \( k \) (for transitions of the second type). Hence, the sum of the next two symmetric numbers (which are closer to the center of the chain) is again congruent either to 1 modulo \( k-1 \) or to 0 modulo \( k \). But the sum of the numbers themselves is not less than 2 and not greater than \( 2k-2 \), therefore it can only be equal to \( k \).

Assume that \( m \) is even, and consider two cases.

1) \( k \) is odd. Then the central transition in the chain is of the second type. The lower-right cell of the diagonal has an odd number, since the number \( n-m \) is odd and \( k-1 \) is even. The upper-left cell of the diagonal also has an odd number; therefore, in a transition of the first type, the parity of the number changes. Suppose from the number 1 a transition of the first type goes to the number \( 2s \). Then modulo \( k-1 \) the transitions of the first type look like \( 1 \rightarrow 2s, 2 \rightarrow 2s+1, \ldots, k-1 \rightarrow 2s+k-2 \). The sums of the numbers in these pairs are consecutive odd numbers, therefore modulo \( k-1 \) they give all odd residues twice. In particular, there is a transition whose sum is 1 modulo \( k-1 \). As shown above, this sum is equal to \( k \). But then the symmetric transition also has the first type and contains the same numbers, which means that one of the transitions is repeated, which is impossible.

2) \( k \) is even. Then the central transition in the chain is of the first type. The last left cell has an odd number, since \( m-1 \) is odd and \( k \) is even. The first right cell also has an odd number, hence under transitions of the second type the parity of the number does not change. Similarly to the first case, one can show that among them there is a transition whose pair of numbers sums to \( k \), giving the same contradiction.

Remark. After describing the order in which the token passes through the cells of the diagonal (from the left it shifts \( k \) cells to the right, and from the right — \( k-1 \) cells), the solution can be completed in a different way.

Number all the cells of the diagonal by the numbers from 1 to \( n \) from left to right. Draw an arrow from each cell to the cell in which the token appears next; these arrows form a path starting at the cell \( k \) and ending at the cell \( m \). Add an arrow leading from the cell \( m \) to the cell \( k \); we obtain a cycle that passes through all the cells of the diagonal.

This cycle determines a permutation \( \sigma \) of the numbers \( 1,2,\ldots,n \), where \( \sigma(i) \) is the number of the cell to which the arrow from cell \( i \) leads. This permutation is a cycle on \( n \) elements. Recall that a permutation which is a cycle on \( b \) elements has parity different from the parity of \( b \). Therefore, the permutation \( \sigma \) is even.

On the other hand, \( \sigma \) is obtained as a composition (successive application) of two permutations: \( \tau \), which sends \( x \mapsto x+(k-1) \bmod n \), and \( \theta \), acting as \( k \mapsto(k+1) \mapsto \cdots \mapsto(m+k-1) \mapsto k \). The permutation \( \tau \) consists of several cycles of equal length; therefore, these cycles have odd length, and hence \( \tau \) is even. Thus, \( \theta \) is also even, which exactly means that \( m \) is odd.

Example. Divide the top and bottom horizontals into horizontal dominoes—they will be inside the \( 2 \times 2 \) squares. The remaining rectangle \( 98 \times 100 \) is divided into vertical dominoes—they will not be inside the \( 2 \times 2 \) squares.

Estimate. Consider the squares \( A_{1}, A_{3}, \ldots, A_{99} \) of sizes \( 1 \times 1, 3 \times 3, \ldots, 99 \times 99 \), whose bottom left corner coincides with the bottom left corner of the original \( 100 \times 100 \) square. For each of the squares \( A_{i} (i = 1, 3, 5, \ldots, 99) \), there is a domino \( X_{i} \) crossing its side (since squares of odd area cannot be divided into dominoes). It is easy to see that \( X_{i} \) lies inside a \( 2 \times 2 \) square of the division.

Similarly, considering the squares \( B_{1}, B_{3}, \ldots, B_{99} \) of sizes \( 1 \times 1, 3 \times 3, \ldots, 99 \times 99 \), whose top right corner coincides with the top right corner of the original \( 100 \times 100 \) square, we find another 50 required dominoes \( Y_{j} (j = 1, 3, 5, \ldots, 99) \).

This completes the solution (it is obvious that all dominoes \( X_{1}, X_{3}, \ldots, X_{99}, Y_{1}, Y_{3}, \ldots, Y_{99} \) are distinct).

Note. We provide a scheme of a slightly different proof of the estimate.

Suppose there are no more than 99 dominoes inside the \( 2 \times 2 \) squares.

Draw 50 vertical grid lines \( v_{1}, v_{3}, v_{5}, \ldots, v_{99} \) such that \( v_{i} \) separates \( i \) columns on the left. It is easy to see that any domino intersected by one of the lines \( v_{1}, v_{3}, v_{5}, \ldots, v_{99} \) is suitable for us. Each vertical line intersects an even number of dominoes, as there is an even number of cells to the left of this line. Therefore, among the lines \( v_{1}, v_{3}, v_{5}, \ldots, v_{99} \), there is a line \( v_{i} \) that does not intersect dominoes, otherwise we have already found at least \( 2 \cdot 50 = 100 \) required dominoes. Conduct a similar reasoning for 50 horizontal grid lines \( h_{1}, h_{3}, h_{5}, \ldots, h_{99} \) and find among them a line \( h_{j} \) that does not intersect dominoes. But \( v_{i} \) and \( h_{j} \) divide the board into regions with an odd number of cells, so at least one of these two lines must intersect a domino. Contradiction.

Associate with the row \(x_{1}, x_{2}, \ldots, x_{100}\) of numbers from 1 to 100 a sequence of 99 signs \(<\) and \(>\) according to the order of adjacent numbers. That is, if \(x_{k} < x_{k + 1}\), then the \(k\)-th sign in this sequence is \(<\), otherwise it is \(>\). Note that the allowed operations on the row do not change the corresponding sequence of signs. Indeed, from the pair of numbers \(x_{k}\) and \(x_{k + 1}\), at most one changes and by at most 1. Therefore, the inequality sign between them cannot change to the opposite. Associate each permutation of signs with an arrangement of numbers according to the following rule (such arrangements will be called distinguished). For \(k = 1,2, \ldots, 99\), if \(x_{k} > x_{k + 1}\) (i.e., the \(k\)-th sign is \(<\)), place the largest of the previously unchosen numbers at \(x_{k}\), otherwise place the smallest of the previously unchosen numbers. Note that with such a sequence of operations, the numbers not chosen in the first \(k\) steps will form a segment of the natural series (and the chosen ones will be several largest and several smallest numbers from 1 to 100). In particular, \(x_{1} = 1\) or \(x_{1} = 100\). Replace the number \(x_{100}\) with the only remaining number.

It is easy to see that the resulting arrangement of numbers corresponds to the chosen sequence of signs. There will be as many distinguished arrangements as there are different sequences of signs, that is, \(2^{99}\). As stated above, no two of the distinguished rows can be made identical by the allowed operations. Thus, by filling the table \(100 \times 2^{99}\) with distinguished rows, we obtain an example for \(n = 2^{99}\).

Now we will prove that from any row \(A\) of length 100, a distinguished row \(B\) with the same arrangement of signs can be obtained. From this, it follows that \(n \leqslant 2^{99}\). Suppose that the first \(t - 1\) signs of the given row \(A = \left(x_{1}, \ldots, x_{100}\right)\) and the distinguished row \(B = \left(y_{1}, \ldots, y_{100}\right)\) coincide. If \(t = 100\), then the rows themselves coincide. Let \(t < 100\). Without loss of generality, assume that \(x_{t} < x_{t + 1}\). As stated above, the sets of numbers \(y_{t}, \ldots, y_{100}\) and \(x_{t}, \ldots, x_{100}\) coincide and form a segment of the natural series with the smallest number \(x_{t}\). Since \(y_{t} < y_{t + 1}\), it is possible to swap in row \(A\) at position \(t\) with a number one less until the number \(x_{t}\) appears at position \(t\). Thus, we have achieved the coincidence of the first \(t\) symbols of our row with the distinguished row \(B\). Therefore, such operations can transform any row into a distinguished row, completing the proof of the estimate.

Let the intersections of line \( MH \) with lines \( AB, AC, BI \), and \( CI \) be denoted by \( P, Q, X \), and \( Y \) respectively (see image 6). Let the lines \( AI, BI \), and \( CI \) intersect \( \omega \) again at points \( A_{1}, B_{1} \), and \( C_{1} \) respectively. Denote \( \angle BAI = \angle CAI = \alpha, \angle ABI = \angle CBI = \beta, \angle ACI = \angle BCI = \gamma \), then \( \alpha + \beta + \gamma = 90^{\circ} \) from the angle sum of triangle \( ABC \). Since \( MH \perp AI \), we have \( \angle AQM = 90^{\circ}-\alpha \). Since quadrilateral \( ABA_{1}C \) is cyclic, \( \angle MA_{1}C = 90^{\circ}-\angle BCA_{1} = 90^{\circ}-\alpha \). Thus, \( \angle MA_{1}C + \angle MQC = 180^{\circ} \), so quadrilateral \( A_{1}CQM \) is cyclic. Therefore, \( \angle AQA_{1} = 90^{\circ} \). Similarly, \( \angle APA_{1} = 90^{\circ} \), from which it follows that points \( A, A_{1}, P, Q \) lie on the circle \( \gamma \), constructed on segment \( AA_{1} \) as a diameter.

Now note that \( \angle QC'C = \angle QYC = 90^{\circ}-\angle CIA_{1} = 90^{\circ}-\alpha-\gamma = \beta \). However, from the cyclicity of quadrilateral \( BC'C B_{1} \), we get that \( \angle CC'B_{1} = \angle B_{1}BC = \beta = \angle CC'Q \). Therefore, points \( C', Q \), and \( B_{1} \) lie on the same line. Similarly, points \( P, B' \), and \( C_{1} \) lie on the same line.

Image 6

Due to the above and the cyclicity of quadrilateral \( C_{1}B_{1}CB \), we have that \( \angle CC_{1}B_{1} = \beta = \angle CYQ \), so \( C_{1}B_{1} \parallel PQ \). Since quadrilateral \( B'C_{1}B_{1}C' \) is cyclic, \( \angle PB'C' = \angle C_{1}B_{1}C' = \angle PQC' \). Thus, quadrilateral \( B'QC'P \) is cyclic. Then the radical axes of its circumcircle, circle \( \gamma \), and circle \( \omega \) intersect at one point, which are the lines \( B'C', PQ \), and \( AA_{1} \). Therefore, point \( H \) lies on line \( B'C' \), which was to be proven.

Remark. We provide another way to complete the solution after establishing that points \( C', Q \), and \( B_{1} \) lie on the same line, and points \( P, B' \), and \( C_{1} \) lie on the same line. Let \( N \) be the midpoint of arc \( BAC \). Let line \( AX \) intersect circle \( \omega \) again at point \( T \). Note that \( \angle CC'Y = \angle AQY = 90^{\circ}-\alpha = \frac{1}{2} \angle CC'B \). Therefore, \( C'Y \) is the angle bisector of angle \( CC'B \), so point \( N \) lies on line \( C'Y \). Similarly, it lies on line \( XB' \). Applying Pascal's theorem to points \( ATC'B_{1}BC \), we obtain that point \( X \), point \( Q \), and the intersection point of \( C'T \) and \( BC \) lie on the same line. Therefore, lines \( BC, XQ \), and \( C'T \) intersect at one point, meaning point \( M \) lies on \( C'T \). Now apply Pascal's theorem to points \( ATC'B'NA_{1} \) and obtain that points \( X \) and \( M \) together with the intersection point of \( AA_{1} \) and \( B'C' \) lie on the same line. Thus, point \( H \) lies on \( B'C' \), which was to be proven.

A polynomial \( P(x) \) is called integer-valued if \( P(k) \) is an integer for any integer \( k \). We need to prove that if the polynomials \( P(x) \) and \( P^{\prime}(x) \) are integer-valued, and the degree of \( P(x) \) is \( d \), then the leading coefficient of the polynomial \( N_{d} \cdot P(x) \) is an integer.

Lemma. Let \( P(x) \) be an integer-valued polynomial of degree \( d \). Then all coefficients of the polynomial \( d!\cdot P(x) \) are integers.

Proof. Consider the polynomial

\[ Q(x) = \sum_{i = 0}^{n} P(i) \cdot \frac{(x - 0)(x - 1) \ldots(x-(i - 1))(x-(i + 1)) \ldots(x - d)}{(i - 0)(i - 1) \ldots(i-(i - 1))(i-(i + 1)) \ldots(i - d)} \]

Its degree is not greater than \( d \), and its values coincide with the corresponding values of \( P(x) \) at the points \( x = 0,1,2, \ldots, d \). This means that the polynomial \( P(x)-Q(x) \) has a degree not higher than \( d \), and also vanishes at \( d + 1 \) points. Therefore, it is zero, i.e., \( P(x) = Q(x) \). (The formula above is a special case of the Lagrange interpolation formula.)

It remains to note that in the formula above, in the \( i \)-th term, the denominator is equal to \( (-1)^{d - i} i!(d - i)!; \) this number divides \( d! \), since \( \frac{d!}{i!(d - i)!} = C_{d}^{i} \). Thus, multiplying each term by \( d! \) results in a polynomial with integer coefficients.

Let's proceed to the solution of the problem. Induction on \( d \). The base case for \( d = 0 \) is trivial. For the induction step, consider a bicelovalued polynomial \( P(x) \) of degree \( d \); let its leading coefficient be \( a \).

If \( d \) is not a prime number, then \( N_{d} = d N_{d - 1} \). Note that the polynomial \( \Delta P(x) = P(x + 1)-P(x) \) is also bicelovalued, has degree \( d - 1 \), and leading coefficient \( a d \). By the induction hypothesis, the number \( N_{d - 1} \cdot a d = N_{d} a \) is an integer, which is what we needed to prove.

Now suppose \( d \) is a prime number; then \( N_{d} = N_{d - 1} \), and the same reasoning shows that the number \( d N_{d} a \) is an integer. Assume that \( N_{d} a \) is not an integer; then the denominator of the number \( a \) (in its irreducible form) is divisible by the prime number \( d \).

Note that the sum of all coefficients of the polynomial \( P(x) \) is the integer \( P(1) \). Since the denominator of the number \( a \) is divisible by \( d \), there will be another coefficient among the coefficients of the polynomial \( P(x) \) whose denominator is divisible by \( d \); let this be the coefficient \( b \) at \( x^{i}, i < d \). Note that \( i > 0 \), since the number \( P(0) \) is an integer.

But then, for the integer-valued polynomial \( P^{\prime}(x) \), the coefficient at \( x^{i - 1} \) is \( i b \) and also has a denominator divisible by \( d \). Since \( d \) is a prime number, it follows that the coefficient at \( x^{i - 1} \) in the polynomial \( (d - 1)!P^{\prime}(x) \) is not an integer, which contradicts the lemma.

Remark 1. The case of prime \( d \) can also be handled in other ways. Here is one of them.

Assume that the number \( d N_{d} a \) is an integer, but \( N_{d} a \) is not, so that \( N_{d} a = t / d \) for some integer \( t \), not divisible by \( d \). Consider the polynomial

\[ Q(x) = N_{d} P(x)-N_{d} a \cdot(x - 1)(x - 2) \ldots(x - d) . \]

It is integer-valued, since for any integer \( k \), the number \( (k - 1)(k- -2) \ldots(k - d) \) is divisible by \( d! \). Moreover, its degree is less than \( d \). From the lemma, it follows that the denominators of the coefficients of the polynomial \( Q(x) \) are not divisible by \( d \).

Now consider the integer \( c = N_{d} P^{\prime}(d) \). From formula (*) it is easy to obtain that

\[ c = Q^{\prime}(d) + \frac{t}{d} \cdot(d - 1)(d - 2) \ldots(d-(d - 1)) . \]

Here, the first term is an irreducible fraction with a denominator not divisible by \( d \), and the second term is divisible by \( d \). This is impossible for an integer \( c \).

Remark 2. As proved by D. Brizolis and E. G. Straus, the smallest \( N \) for which the leading coefficient of the polynomial \( N P(x) \) is necessarily an integer is equal to \( d!\prod_{p} p^{-k(p, d)} \), where the product is taken over all primes \( p \), and \( k(p, d) \) is the largest non-negative integer \( k \) for which the inequality \( k p^{k}-(k - 1) p^{k - 1} \leqslant d \) holds.

If you completely remove 51 piles, then obviously there will not be many stones left. Therefore, the desired value of \( n \) is less than 5100. (Alternatively, you can remove 51 stones from each pile.)

It remains to show that after removing any \( n = 5099 \) stones, there will still be many stones. Let there be \( a_{1}, a_{2}, \ldots, a_{100} \) stones left in the piles, respectively; we can assume that \( 0 \leqslant a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{100} \leqslant 100 \). We will show that \( a_{i + 50} \geqslant i \) for \( i = 1,2, \ldots, 50 \), meaning that the piles numbered from 51 to 100 meet the requirements.

Suppose this is not the case, that is, \( a_{i + 50} \leqslant i - 1 \) for some \( i \leqslant 50 \). This means that each of the first \( i + 50 \) piles contains no more than \( i - 1 \) stones, meaning at least \( 101-i \) stones have been removed from it. Therefore, the total number of removed stones is at least \( (i + 50)(101-i) = 5100-(i - 1)(i - 50) \geqslant 5100 \). Contradiction.

To each remainder \( a \) modulo 19, associate a remainder \( b(a) \) such that \( b(a) \equiv 3a \pmod{19} \). Note that the remainders \( 0, 1, 2, 3, 7, 8, 9 \) correspond to the remainders \( 0, 3, 6, 9, 2, 5, 8 \) respectively. Moreover, from the remainder \( b \), we can recover the remainder \( a = a(b) \equiv -6b \pmod{19} \) such that \( a(b(a)) \equiv -18a \equiv a \pmod{19} \) and \( b(a(b)) = b \) (by similar reasoning).

Now, let \( \mathcal{A} \) be the set of numbers from the condition that do not contain the digits \( 4, 5, 6 \), and let \( \mathcal{B} \) be the set of such numbers that do not contain \( 1, 4, 7 \). To each number \( A = \overline{a_{99} a_{98} \ldots a_{0}} \in \mathcal{A} \), associate the number \( B = \overline{b(a_{99}) b(a_{98}) \ldots b(a_{0})} \). Note that \( b(a_{i}) \) is a digit (and \( b(a_{99}) \neq 0 \)), so we obtain a 100-digit number. Furthermore,

\[ B = b_{0} + 10b_{1} + \ldots + 10^{99}b_{99} \equiv 3(a_{0} + 10a_{1} + \ldots + 10^{99}a_{99}) = 3A \pmod{19} \]

so \( B \) is divisible by 19 and \( B \in \mathcal{B} \). Since different numbers from \( \mathcal{A} \) correspond to different numbers from \( \mathcal{B} \), the number of numbers in \( \mathcal{B} \) is not less than in \( \mathcal{A} \).

Finally, to each number \( B = \overline{b_{99} b_{98} \ldots b_{0}} \in \mathcal{B} \), there corresponds a number \( A = \overline{a(b_{99}) a(b_{98}) \ldots a(b_{0})} \), which for similar reasons lies in \( \mathcal{A} \). It follows that the numbers in \( \mathcal{A} \) and \( \mathcal{B} \) are equal.

We will show that Petya can determine the weight of one weight, even if he has 8,000 weights. Let \( n = 4000 \).

Lemma. For any \( n \) weights, Petya can find two weights for which he knows their total weight.

Proof. Let Petya sequentially put all possible sets of 10 weights from our \( n \) into the device. Note that each reading of the device is the weight of some pair of weights from \( C_{n}^{2} \) (we will say that this reading uses this pair). At the same time, Petya will get \( C_{n}^{10} \) readings. Therefore, one of the pairs will be used at least

\[ D = \frac{C_{n}^{10}}{C_{n}^{2}} = \frac{(n - 2)(n - 3) \ldots(n - 9)}{3 \cdot 4 \cdot \ldots \cdot 10} \]

times.

In other words, there will be \( D \) measurements such that (1) the device shows the same weight \( S \), and (2) in all tens used in these tests, there are two common weights \( a \) and \( b \). We will show that if conditions (1) and (2) are met, the total weight of \( a \) and \( b \) must be equal to \( S \), that is, Petya will be able to determine the weight of this pair from the device readings. We will call the tens of weights involved in these \( D \) measurements necessary.

Assume the contrary: the sum of the weights \( a \) and \( b \) is not equal to \( S \). Consider all pairs from \( n \) weights whose total weights are equal to \( S \) (we will call these pairs good). Since the weights of all weights are different, good pairs do not intersect; in particular, there are no more than \( n / 2 \) of them. At the same time, each necessary ten contains not only weights \( a \) and \( b \), but also at least one good pair. Now estimate the total number of necessary tens.

Let in a necessary ten the good pair not contain either \( a \) or \( b \). Any such ten can be obtained by adding a good pair to weights \( a \) and \( b \) (in no more than \( (n - 2) / 2 \) ways), and then completing with six of the remaining \( n - 4 \) weights. In total, the number of such tens is no more than \( \frac{n - 2}{2} C_{n - 4}^{6} \).

In all other necessary tens, the good pair contains either \( a \) or \( b \). If there is a good pair containing \( a \), then such a pair is unique. To obtain a necessary ten containing this pair, it must be supplemented with weight \( b \) and seven more weights from the remaining \( n - 3 \); in total, there are no more than \( C_{n - 3}^{7} \) such necessary tens. Similarly, there are no more than \( C_{n - 3}^{7} \) necessary tens containing a good pair with weight \( b \).

In total, we get

\[ \begin{aligned} D \leqslant \frac{n - 2}{2} & C_{n - 4}^{6} + 2 C_{n - 3}^{7} = \\ & \quad = D \cdot\left(\frac{7 \cdot 8 \cdot 9 \cdot 10}{4(n - 3)} + \frac{8 \cdot 9 \cdot 10}{n - 2}\right) < D \cdot\left(\frac{1}{2} + \frac{1}{2}\right) = D \end{aligned} \]

Contradiction.

We will complete the solution of the problem. Construct the following graph. Assign each weight a vertex. Among every \( n \) weights, find one pair with a known sum; connect the two corresponding vertices with an edge. If there are no odd cycles in this graph, then, as is known, its vertices can be colored in two colors so that each edge connects vertices of different colors. But then the vertices of one color are no less than \( n \), and therefore we have drawn an edge among them; contradiction.

Thus, there is a cycle \( w_{1}, w_{2}, \ldots, w_{2 k + 1} \) in the resulting graph, and Petya knows the total weights of all pairs of neighboring weights in this cycle. By taking the half-sum of all these weights, Petya will find out the total weight of all weights in the cycle. Subtracting from it \( \left(w_{2} + w_{3}\right) + \left(w_{4} + w_{5}\right) + \ldots + + \left(w_{2 k} + w_{2 k + 1}\right) \), he will find out the weight of weight \( w_{1} \).

Note. By estimating a little more precisely, the lemma can be proven even for \( n = 2000 \).

Let the line \(EC\) intersect the circumcircle \((ABC)\) again at point \(X\), and the line \(EA\) intersect the circumcircle \((ACD)\) again at point \(Y\) (we consider the arrangement of points as shown in image 5; other cases are considered similarly).

Consider the homothety with center \(E\) that maps \((ABC)\) to \((ACD)\). Under this homothety, point \(X\) maps to \(C\), and point \(A\) maps to \(Y\). Hence, \(AX \parallel YC\) and \(\angle AEC = \angle AYC - \angle ECY = \angle AYC - \angle AXC\).

Image 5

But \(\angle AXC = 180^\circ - \angle ABC\) and \(\angle AYC = 180^\circ - \angle ADC\). Therefore, \(\angle AEC = \angle ABC - \angle ADC = \angle ABC - \angle BCG = \angle BGC\). From this equality, it follows that the points \(A, C, E, G\) lie on a circle.

Since point \(E\) lies on the perpendicular bisector of \(AC\) (i.e., on the axis of symmetry of the circles \((ABC)\) and \((ACD)\)), it is the midpoint of the arc \(AGC\) of the circle \((ACEG)\). Thus, \(E\) lies on the external bisector of angle \(BGC\).

Similarly, it can be shown that \(F\) also lies on the external bisector of angle \(BGC\).

Remark. The problem has the following generalization. Let \(ABCD\) be a quadrilateral, \(G = AB \cap CD\), and \(M\) be the second intersection point of the circles \((ADG)\) and \((BCG)\) (in other words, the Miquel point of this quadrilateral). Let \(E\) be the center of the homothety with a positive coefficient that maps \((ABC)\) to \((ADC)\). Then the points \(A, C, M, E\) lie on a circle, and \(E\) is the midpoint of the arc \(AC\) (i.e., \(ME\) is the bisector of the angle between \(AM\) and \(CM\)).

This can be proven similarly to the solution of the problem: we have (in directed angles) \(\angle AEC = \angle ABC + \angle ADC = \angle GBC + \angle AMG = \angle GMC + \angle AMG = \angle AMC\).

Note that equality is achieved when \( x_{0} = a \) and \( x_{1} = \ldots = x_{n} = 1 \).

We write \( \sum x_{k}^{2} = \sum\left(1-x_{k}\right)^{2} + 2 \sum x_{k}-(n + 1) = \sum\left(1-x_{k}\right)^{2} + \)

\( + n - 1 + 2 a \). It is sufficient to prove that \( \sum\left(1-x_{k}\right)^{2} \geqslant(1-a)^{2} \). Let \( x_{0} \) be the smallest of the numbers.

If \( x_{0} \leqslant a \), we have \( \sum\left(1-x_{k}\right)^{2} \geqslant\left(1-x_{0}\right)^{2} \geqslant(1-a)^{2} \).

If \( x_{0} \geqslant a \), then \( \sum\left(1-x_{k}\right)^{2} = \sum x_{k}\left(\frac{1}{x_{k}}-2 + x_{k}\right) \), which, since the expressions in the brackets are non-negative, is not less than \( a \sum\left(\frac{1}{x_{k}}-2 + x_{k}\right) = a\left(n + \frac{1}{a}-2(n + 1) + n + a\right) = = a\left(\frac{1}{a}-2 + a\right) = (a - 1)^{2} \).

Note that \( 2023 = 7 \cdot 17^{2} \). Grisha will divide the numbers on the board into two groups of 5 and will square the numbers from the first group and the second group alternately. It is easy to see that the squares of integers not divisible by 7 can only give remainders of 1, 2, and 4 when divided by 7. Therefore, after increasing by at most 2, the numbers on the board will give remainders of only \( 1, 2, 3, 4, 5 \), and 6 when divided by 7. Thus, none of the numbers will be divisible by 7, and therefore not by 2023.

Note: There are other solutions.

The networks considered with \( N - k \) roads are further referred to as \( k \)-networks. Consider the undirected graph formed by the roads of a \( k \)-network. It has at most \( k \) connected components, since each contains a selected city. On the other hand, there are at least \( k \) components, since there are at most \( N - k \) edges. Therefore, there are exactly \( k \) components, each of which is a tree, contains a single selected city, and—recalling the orientation—the edges of each tree are directed towards the selected city. In particular, from each non-selected city, exactly one road must exit, and from a selected city, 0 roads.

Consider a \( (k + 1) \)-optimal network \( A \) with selected cities \( y_{0}, y_{1}, \ldots, y_{k} \) and a \( k \)-optimal network \( B \) with selected cities \( x_{1}, \ldots, x_{k} \). Without loss of generality, none of the \( x_{i} \) can reach city \( y_{0} \) in network \( A \). Let \( U \) be the set of cities from which \( y_{0} \) can be reached in \( A \), and let \( \alpha, \beta \) be the sets of roads exiting \( U \) in networks \( A, B \) respectively. We have \( |\alpha| = |U|-1,|\beta| = |U| \).

Consider the network \( D : = (A \backslash \alpha) \cup \beta \). It is claimed that this is a \( k \)-network for the selected cities \( y_{1}, \ldots, y_{k} \). Indeed, the number of roads in it is \( |D| = N - k - 1-(|U|-1) + |U| = N - k \). From each city, except \( y_{1}, \ldots, y_{k} \), exactly one road exits. Leaving any city outside \( U \) and using the network roads, we can still reach one of the cities \( y_{1}, \ldots, y_{k} \). Leaving a city in \( U \), we either exit \( U \) and then reach one of the cities \( y_{1}, \ldots, y_{k} \), or cycle within \( U \). But then \( \beta \) contains a cycle, which is impossible.

Consider the network \( C : = (B \backslash \beta) \cup \alpha \). It is claimed that this is a \( (k + 1) \)-network for the selected cities \( x_{1}, \ldots, x_{k}, y_{0} \). Indeed, \( |C| = n - k - 1 \), and leaving any city via the roads of network \( C \), we either enter \( U \) and then reach \( y_{0} \) via \( \alpha \), or never enter \( U \) and then reach one of the cities \( x_{1}, \ldots, x_{k} \).

Thus, \( C, D \) are \( k \)-network and \( (k + 1) \)-network. The sum of their costs is the same as that of \( A \) and \( B \). Hence, they are both optimal. Thus, for network \( A \), it was possible to remove the selected city and find an optimal \( k \)-network with the remaining selected cities. Now, the required numbering can be constructed in reverse order (starting with an empty \( N \)-network).