Example: \( 1,2,3,4,5,6,8,10,12,20 \). In this set, three numbers \( (5,10,20) \) are divisible by 5, four numbers \( (4,8,12,20) \) are divisible by 4, and the total sum is 71.

Note. It can be proven (though, of course, this is not required in the problem) that in any example satisfying the problem's condition, the numbers \( 1,2,3,4,5,8,10,12 \) and 20 must necessarily be present, and instead of the number 6, one can take 7 or 9.

Let \( S \) be the sum of the digits of the number \( N \). Then the sums of the digits of the resulting numbers will be \( S + 1, S + 2, \ldots, S + 9 \). Three of these sums will be divisible by 3. By the divisibility rule for 3, the corresponding three numbers on the board will also be divisible by 3. Since they will be greater than 3, they will be composite. Therefore, there cannot be more than 6 prime numbers on the board.

Six prime numbers can indeed appear, for example, if \( N = 1 \) - Petya could obtain, among others, the numbers \( 11, 13, 41, 61, 17, \) and \( 19 \).

Note. Petya can obtain six prime numbers even if he appends digits only on one side. For example, from the number \( N = 3 \), he can obtain the numbers \( 13, 23, 43, 53, 73, \) and \( 83 \).

Let \( d \) be the abscissa of the vertex of the parabola \( y = P(x) \), so that the line \( x = d \) is the axis of symmetry of the parabola. Then for any numbers \( t \) and \( s \) with the sum \( 2d \) (i.e., such that the points \( t \) and \( s \) are symmetric with respect to \( d \)), it holds that \( P(t) = P(s) \). Thus, any triplet of pairwise distinct numbers \( a, b, c \) such that \( a + b + c = 2d \) will satisfy the condition of the problem. For example, we can take \( a = \frac{2d}{3} - 1, b = \frac{2d}{3}, c = \frac{2d}{3} + 1 \).

Let \( P(x) = u x^{2} + v x + w \). According to the condition,

\[ \begin{aligned} (a - b)(u(a + b) + v) = \left(u a^{2} + v a + w\right)-\left(u b^{2}\right. & + v b + w) = \\ & = P(a)-P(b) = n^{2} \end{aligned} \]

Since \( n \) is a natural number, \( a - b \neq 0 \). We will look for suitable pairs of numbers \( (c, d) \) in the form \( c = a + k \) and \( d = b - k \). Then the difference \( P(c)-P(d) \) will be

\[ (c - d)(u(c + d) + v) = (a - b + 2 k)(u(a + b) + v) = \frac{a - b + 2 k}{a - b} n^{2} . \]

Thus, we need to choose such \( k \) for which the fraction \( \frac{a - b + 2 k}{a - b} \) will be a square of a natural number. For \( k = (a - b) m \), the fraction will be an odd number \( 2 m + 1 \). Then we need those \( m \) for which \( 2 m + 1 \) will be a square of an odd number.

Note. Notice that for the polynomial in the condition, some (or even all!) coefficients may turn out to be irrational. For example, the condition is satisfied for the polynomial \( P(x) = u x^{2} + (1-u) x - u \) and numbers \( a = -1, b = 2 \), where \( u \) is an arbitrary number.

Note that if the number \( u \) is irrational, then the expression \( P(c)-P(d) \) can only be an integer when \( c + d = a + b \) or \( c = d \).

In all the solutions below, we consider the friendship graph, where vertices are people in the company, and two people are connected by an edge if they are friends.

If the graph is a cycle containing 101 vertices, then on any 100 vertices there are exactly 99 edges, so such a company satisfies the conditions of the problem. It remains to show that there cannot be such a company with 102 people (then a company with more than 102 people also cannot exist). Below we provide several different ways to do this; in each method, we assume, for contradiction, that such a company exists.

First solution. Consider an arbitrary set of 101 vertices and the induced subgraph on these vertices, let there be \( k \) edges. By removing an arbitrary vertex (say, of degree \( d \)) from this subgraph, we get 100 vertices with an odd number of edges \( k - d \). Thus, the degree of any vertex in our subgraph has a parity different from the parity of \( k \), that is, the degrees of all vertices in the subgraph have the same parity. But, as is well known, in a graph with an odd number of vertices, all vertices cannot have odd degrees. Therefore, all these degrees are even, and the number of edges \( k \) is odd.

Now let there be \( \ell \) edges in the entire graph on 102 vertices. When removing any vertex (say, of degree \( d \)), we get a subgraph with an odd number of edges \( \ell - d \); similarly to the reasoning above, we conclude that in the entire graph, the degrees of all vertices have the same parity. Note that our graph cannot be empty (i.e., have no edges) or complete (i.e., have all \( C_{102}^{2} \) edges), otherwise on any 100 vertices there would be either 0 or \( C_{100}^{2} = 99 \cdot 50 \) edges, that is, an even number. Then there will be a vertex connected to at least some other vertex, but not to all. In other words, there are vertices \( u, v_{1} \), and \( v_{2} \) such that \( u \) is connected to \( v_{1} \) but not to \( v_{2} \). The degrees of vertices \( v_{1} \) and \( v_{2} \) in the original graph have the same parity, so after removing \( u \), they will have different parities. This is impossible as proven above.

Second solution. There are a total of \( n = C_{102}^{2} = 51 \cdot 101 \) ways to remove two vertices from 102, leaving 100. Number these ways from 1 to \( n \). Let \( a_{i} \) be the number of edges on the remaining 100 vertices in the \( i \)-th way; by assumption, all numbers \( a_{i} \) are odd, and therefore their sum \( S \) is odd (since the number \( n \) is odd).

On the other hand, consider any edge uv. This edge is counted in the number \( a_{i} \) exactly when vertices \( u \) and \( v \) are not removed in the \( i \)-th way, that is, when some pair of the remaining 100 vertices is removed. This happens in \( k = C_{100}^{2} = 50 \cdot 99 \) ways. Thus, each edge is counted in \( S \) an even number \( k \) of times, so \( S \) must be even. Contradiction.

Third solution. Call a vertex even if its degree is even, and odd otherwise. Consider two cases.

Case 1. Suppose the total number of edges in the graph is odd. Then, by removing any pair of vertices, we must remove an even number of edges (so that an odd number remains). On the other hand, if we remove vertices with degrees \( d_{1} \) and \( d_{2} \), then the number of removed edges is \( d_{1} + d_{2} \) if these vertices are not connected by an edge, and \( d_{1} + d_{2}-1 \) if they are connected. It follows that vertices of the same parity are always not connected by an edge, and vertices of different parities are always connected.

Thus, if there are \( k \) even vertices and \( \ell \) odd vertices in the graph, then even vertices have (even) degree \( \ell \), and odd vertices have (odd) degree \( k \). This is impossible, because \( k + \ell = 102 \).

Case 2. Suppose the total number of edges in the graph is even. Similarly, we obtain that vertices of the same parity are always connected by an edge, and vertices of different parities are not connected. Therefore, if there are \( k \) even vertices and \( \ell \) odd vertices in the graph, then even vertices have (even) degree \( k - 1 \), and odd vertices have (odd) degree \( \ell - 1 \). This again contradicts the equality \( k + \ell = 102 \).

Note. Of course, there are other examples of a company of 101 people that satisfy the condition.

In all solutions below, the circle circumscribed around triangle \( CEF \) is denoted by \( \omega \), and its center by \( O \).

First solution. The external (\( CD \)) and internal (\( CE \)) bisectors of angle \( ACB \) are perpendicular. Then \( \angle DAE = \angle DCE = 90^{\circ} \), and quadrilateral \( ADCE \) is inscribed. Therefore, \( \angle ADE = \angle ACE \), from which \( \angle BEF = 180^{\circ} - \angle AED - \angle DEF = 90^{\circ} - \angle AED = \angle ADE = \angle ACE = \angle ECF \). Thus, \( \omega \) is tangent to \( AB \) (at point \( E \); see fig. 1).

Choose a point \( P \) on segment \( BC \) such that \( PE \parallel AC \). Then \( \angle PEC = \angle ECA = \angle PCE \), from which \( PC = PE \). Since \( OC = OE \), we obtain that \( PO \) is the bisector of the external angle \( BPE \).

Perform a homothety with center \( B \) that maps triangle \( BPE \) to \( BCA \). Then line \( PO \) will map to \( CD \), and line \( EO \) (perpendicular to \( AB \)) will map to \( AD \). Thus, point \( O \) will map to \( D \), from which it follows that point \( O \) lies on \( BD \).

Fig. 1

Fig. 2

Second solution. Choose points \( D' \) and \( F' \) on lines \( CD \) and \( AC \) such that \( \angle ABD' = \angle D'EF' = 90^{\circ} \) (see fig. 2). As above, we prove that circle \( \omega \) is tangent to \( AB \). Similarly, the circumscribed circle of triangle \( CEF' \) is also tangent to \( AB \) at point \( E \) (and also passes through \( C \)). Therefore, these circles coincide, meaning \( F' \) lies on \( \omega \).

We will prove that the center of \( \omega \) is point \( T \), the intersection of the diagonals \( BD \) and \( AD' \) of trapezoid \( ADD'B \). Note immediately that from the inscribed quadrilaterals \( ADCE \) and \( BD'CE \), we obtain the equalities \( \angle AED = \angle ACD = \angle BCD' = \angle BED' \), so the right triangles \( AED \) and \( BED' \) are similar, and \( \frac{AE}{BE} = \frac{AD}{BD'} = \frac{DT}{BT} \). Therefore, \( ET \parallel AD \parallel BD' \).

Let \( TE \) intersect \( DD' \) at point \( E' \). Since \( E' \) lies on the diameter of \( \omega \) passing through \( E \), and \( \angle ECE' = 90^{\circ} \), point \( E' \) lies on \( \omega \). Line \( EE' \) is parallel to the bases of the trapezoid and passes through the intersection point of the diagonals of the trapezoid; as is known, in this case, \( EE' \) is bisected by point \( T \). Since \( EE' \) is the diameter of \( \omega \), we obtain \( T = O \).

Remark. After proving that \( \omega \) is tangent to \( AB \) at point \( E \), there are other ways to continue the solution. Here is a sketch of one of them.

It is easy to see that \( \angle CEF = \angle CAE = \mu \). Mark a point \( X \) on ray \( BD \) such that \( \angle BCX = 90^{\circ} - \angle CEF = \angle DAC \); then \( \angle XCD = \angle CDA = \nu \). Applying the sine theorem, we can obtain that

\[ \frac{BX}{XD} = \frac{BC}{CD} \cdot \frac{\cos \mu}{\sin \nu} = \frac{BC}{CD} \cdot \frac{CD}{CA} = \frac{BC}{CA} = \frac{BE}{EA}. \]

Thus, point \( X \) lies on lines \( EO \) and \( CO \), that is, \( O = E \) (if these two lines do not coincide, that is, \( AC \neq BC \)).

By writing the condition for \( n = 2022, k = 1011 \) and for \( n = 1011, k = 2022 \), we obtain

\[ a_{2022} = a_{2022} - a_{1011} \geqslant 2022^{3} - 1011^{3} \]

and

\[ -a_{2022} = a_{1011} - a_{2022} \geqslant 1011^{3} - 2022^{3}, \]

which means \( a_{2022} \geqslant 2022^{3} - 1011^{3} \geqslant a_{2022} \). Hence, the answer follows.

Note. A sequence satisfying the condition exists, namely \( a_{n} = n^{3} - 1011^{3} \). Moreover, similarly to the solution above, it is not difficult to show that such a sequence is unique. However, to solve the problem, it is not necessary to find all such sequences or even provide an example of such a sequence.

First solution. Number the columns from left to right with numbers from 1 to 100. Let \( a \) be the top row of the square, and \( b \) be the row immediately below it. Suppose in Petya's division, these rows are occupied by vertical dominoes \( a 1-b 1, a 2-b 2, \ldots, a 98-b 98 \) and horizontal dominoes \( a 99-a 100, b 99-b 100 \). Obviously, the remaining part of the board can be divided into dominoes (for example, into horizontal ones), so such a division exists.

Assume there exists Vasya's division into dominoes that meets the problem's requirements. If in Vasya's division any of the cells \( a 1, a 2, \ldots, a 98 \) is occupied by a vertical domino, then it is the same domino as in Petya's division, and from these two cells, one cannot reach the others. Therefore, in Vasya's division, dominoes \( a 1-a 2, a 3-a 4, \ldots, a 97-a 98 \) must be present. Similarly, cells \( a 99 \) and \( a 100 \) cannot be covered by horizontal dominoes, so they are covered by vertical dominoes \( a 99-b 99 \) and \( a 100-b 100 \). But then from the four cells \( a 99, a 100, b 99, b 100 \), one cannot reach the others—a contradiction.

Note. There are other variants of Petya's division where the required is impossible. For example, if we denote by \( c \) the row directly below \( b \), then any division containing the following five dominoes is suitable: \( a 1-b 1 \), \( a 2-b 2, a 3-a 4, b 3-b 4, c 1-c 2 \) (such divisions exist).

Second solution. Suppose Vasya succeeded in the required task. Then from each cell, one blue and one red segment emerges, and they go to different cells—otherwise, from these two cells, one could not reach the others.

Color all cells in a checkerboard pattern in black and white, and place an arrow on each blue segment from the white cell to the black one, and on the red one—from the black to the white. Then from each cell, exactly one arrow leads out, and exactly one enters it. Then all cells are divided into cycles, and if Vasya succeeded in the required task, then there is one cycle of all cells.

Let \( a \) be the top horizontal line, and \( z \) the bottom one. Suppose in Petya's division there are dominoes \( a 1-a 2 \) and \( z 2-z 3 \) (such a division is possible if, for example, cells \( z 1 \) and \( z 100 \) are covered by vertical dominoes, and all other dominoes are made horizontal). Then these segments will be oriented as \( a 1 \rightarrow a 2 \) and \( z 2 \rightarrow z 3 \). If they are in the same cycle, then this cycle must go from \( a 2 \) to \( z 2 \), and then from \( z 3 \) to \( a 1 \). But such two paths must have a common cell, which is impossible.

Since \(AD \parallel CB\), triangles \(EAD\) and \(ECB\) are similar, and therefore \(\frac{BE}{DE} = \frac{BC}{AD}\).

Extend triangle \(BCD\) to parallelogram \(BCDK\) (see figure ??). Then triangles \(DEF\) and \(DBK\) are similar, so \(\frac{DF}{DE} = \frac{DK}{DB}\). Finally, since \(DK = BC\) and \(DB = DA\), we have

\[ \frac{DF}{DE} = \frac{DK}{DB} = \frac{BC}{AD} = \frac{BE}{DE} \]

from which it follows that \(DF = BE\). Fig. 3

First solution. Example. First, we show that for \( N = 180 \), the requirement is possible. Mark 180 points on a circle, dividing it into 180 equal arcs of \( 2^{\circ} \) each. The measure of any arc with endpoints at two of the marked points is expressed as an even number of degrees, so the measure of any inscribed angle in the circle, formed by three marked points, is expressed as a natural number of degrees. Therefore, the 180 marked points satisfy the condition of the problem.

Estimation. It remains to prove that \( N \leqslant 180 \). Any three marked points form a triangle, so they cannot lie on a straight line. Assuming the marked points are located on the coordinate plane, denote by \( A \) any of them with the maximum ordinate. Among the remaining points, choose points \( B \) and \( C \) such that the angle \( B A C \) is maximal.

From the condition of the problem, it follows that in triangle \( A B C \), the angles \( A B C \) and \( A C B \) are not less than \( 1^{\circ} \), so the angle \( B A C \) is not more than \( 178^{\circ} \). Due to the choice of points \( B \) and \( C \), the remaining \( N - 3 \) marked points lie strictly inside the angle \( B A C \), and each ray starting at point \( A \) contains no more than one of them. By drawing a ray from point \( A \) through each marked point inside the angle \( B A C \), we obtain \( N - 3 \) different rays, dividing \( \angle B A C \) into \( N - 2 \) angles. If \( N - 2 > 178 \), then at least one of these angles has a measure less than \( 1^{\circ} \) and is an angle of some triangle with vertices at three marked points, which contradicts the condition of the problem. Therefore, \( N - 2 \leqslant 178 \), i.e., \( N \leqslant 180 \), which was to be proved.

Remark 1. The choice of the point \( A \) used in the solution can also be described in the following ways.

1. Consider the vertex \( A \) of the convex hull of the system of marked points. As points \( B \) and \( C \), one can then take the vertices of the convex hull adjacent to \( A \).

2. Consider the supporting line of the set of marked points, i.e., a line such that all marked points lie on one side of this line, and at least one marked point lies on the line itself. This point can be taken as point \( A \).

Remark 2. In the example, the marked points are the vertices of a regular 180-gon. All examples for \( N = 180 \) are arranged in this way (this can be easily derived using the reasoning from the proof of the estimation, but of course, this is not required in the solution).

Second solution. We provide another proof of the estimation. Consider a pair of marked points \( A, B \) at the greatest distance from each other. Then for any other marked point \( C \), the side \( A B \) is the largest in triangle \( A B C \), so, in particular, the angle \( \angle B A C \) is acute.

By drawing rays from point \( A \) to all marked points, we find that all these rays are distinct (since three marked points cannot lie on a straight line), and each forms an acute angle with the ray \( A B \), expressed as an integer number of degrees. Such an angle (if the ray does not coincide with \( A B \)) can take values from \( 1^{\circ} \) to \( 89^{\circ} \), so the number of such rays \( N - 2 \) does not exceed \( 2 \cdot 89 = 178 \). Hence \( N \leqslant 180 \).

Remark 3. We can prove weaker estimates \( N \leqslant 361 \) and even \( N \leqslant 181 \) by considering any marked point \( A \) (without any special choice) and the rays emanating from it to other marked points. Indeed, since the angle between any two such rays is measured in whole degrees, the possible directions of these rays are 360, hence \( N \leqslant 361 \). Moreover, from any pair of opposite directions, no more than one can be present, so there are no more than 180 rays, and \( N \leqslant 181 \).

Remark 4. The estimation can be proved somewhat differently. Consider the angle \( B A C \) of the convex hull of the set of marked points. It is expressed as a natural number of degrees and is less than \( 180^{\circ} \), so it does not exceed \( 179^{\circ} \). Repeating the reasoning from the solution, we obtain that \( N - 2 \leqslant 179 \), hence \( N \leqslant 181 \).

If at least one angle of the convex hull is not more than \( 178^{\circ} \), then we prove the estimate \( N \leqslant 180 \) as in the first solution.

The remaining case is when all angles of the convex hull are equal to \( 179^{\circ} \), or all external angles of the convex hull are equal to \( 1^{\circ} \). But the sum of the external angles of a convex polygon is \( 360^{\circ} \), so the convex hull has at least 360 vertices, then \( N \geqslant 360 \). But it was previously shown that \( N \leqslant 181 \) - a contradiction.

Let \( a = 10^{100} \). Denote by \( s(m) \) the sum of the digits of the number \( m \). Note the simple property \( s(\ell) + s(m) \geqslant s(\ell + m) \), which is immediately evident if the numbers \( \ell \) and \( m \) are added column-wise.

Lemma. Let \( k \) be a natural number, and let the natural number \( m \) be divisible by \( 10^{k}-1 \). Then \( s(m) \geqslant 9 k \).

Proof. Induction on \( m \). The base case \( m = 10^{k}-1 \) is obvious.

Assume that \( m \geqslant 10^{k} \), and that the statement is proven for all numbers less than \( m \). We will prove it for \( m \). Let the last \( k \) digits of the number \( m \) form the number \( v \) (possibly with leading zeros), and all the others form the number \( u > 0 \) (in other words, \( m = \overline{u v} = 10^{k} u + v \) ). Since \( m \) is divisible by \( 10^{k}-1 \), the (positive) number \( m^{\prime} = u + v = m - (10^{k}-1) u \) is also divisible by \( 10^{k}-1 \). Therefore, \( s(m^{\prime}) \geqslant 9 k \) by the induction hypothesis, and then \( s(m) = s(u) + s(v) \geqslant s(u + v) = s(m^{\prime}) \geqslant 9 k \).

To solve the problem, it remains to choose such \( k \) that \( 9 k \geqslant a \), and note that if \( b = 10^{k}-1 \) and \( n \geqslant b \), then \( n! \) is divisible by \( 10^{k}-1 \) and, therefore, \( s(n!) \geqslant 9 k \geqslant a \).

Remark. The proof of the lemma essentially uses the following divisibility criterion for \( 10^{k}-1 \). Divide the decimal representation of the number \( m \) into blocks of \( k \) digits (the first block may be incomplete). Treating these blocks as ordinary numbers (possibly with leading zeros), sum them to obtain the number \( m^{\prime} \). Then \( m \) is divisible by \( 10^{k}-1 \) if and only if \( m^{\prime} \) is divisible by \( 10^{k}-1 \).

The lemma has several variations; for example, any number divisible by the number \( \underbrace{11 \ldots 1}_{k \text{ ones}} \) has a digit sum not less than \( k \).

Estimation. Let's prove that \( n > 160 \) does not work.

Suppose there is a set of \( n \) candies. We call a type critical if there are exactly 10 candies of this type (among all \( n \) candies). Let there be \( k \) critical types, then the total number of candies is at least \( 10k : n \geqslant 10k \). Remove one candy of each critical type and organize a group of the remaining \( n - k \) candies. For this group, there is no type represented by exactly 10 candies. Moreover, \( n - k \geqslant n-\frac{n}{10} = \frac{9n}{10} > \frac{9 \cdot 160}{10} = 144 \). Therefore, the considered group contains at least 145 candies, so the condition of the problem is not satisfied.

Example. Now let's give an example of a situation where Vasya can have 160 candies. Suppose he has exactly 10 candies of 16 types. Suppose a group is chosen for which there is no type represented by exactly 10 candies. Then this group does not include at least one candy of each type (in other words, no type is taken completely), i.e., the group contains at most \( 16 \cdot 9 = 144 \) candies, so the condition of the problem is satisfied.

Remark 1. The estimation \( n \leqslant 160 \) can also be proven in the following slightly different way. Suppose some \( n = 145 + m \) "works," i.e., there exists a set \( K_{0} \) of \( n \) candies satisfying the condition of the problem. Then in the set \( K_{0} \) there are exactly 10 candies of some type \( A_{0} \). Remove a candy \( a_{0} \) of type \( A_{0} \) and consider the group \( K_{1} \) of the remaining \( n - 1 \) candies. In the group \( K_{1} \) there are exactly 10 candies of some type \( A_{1} \), while type \( A_{1} \) is different from \( A_{0} \), since there are exactly 9 candies of type \( A_{0} \) in \( K_{1} \). Remove a candy \( a_{1} \) of type \( A_{1} \) from \( K_{1} \) and consider the group \( K_{2} \) of the remaining \( n - 2 \) candies, in which we find 10 candies of a new type \( A_{3} \), and so on, continue step by step removing one candy and considering groups of remaining candies. Thus, we reach the group \( K_{m} \), consisting of \( \left|K_{m}\right| = 145 \) candies. According to our algorithm, \( K_{0} \) contains 10 candies of types \( A_{0}, \ldots, A_{m} \), so \( \left|K_{0}\right| \geqslant 10(m + 1) \). We have \( 145 + m \geqslant 10(m + 1) \), from which \( m \leqslant 15 \) and \( n \leqslant 160 \).

Remark 2. The example for \( n = 160 \) is unique.

It is sufficient to prove the statement: the polynomial \( P(x) \) is divisible by \( (x - 1)^{2} \). Indeed, after division (for example, by long division), the quotient will be a polynomial \( Q(x) \) with integer coefficients, and then the polynomial equality \( P(x) = (x - 1)^{2} Q(x) \) implies the equality \( P(2022) = 2021^{2} \cdot Q(2022) \), from which the statement of the problem follows, since \( Q(2022) \) is an integer.

To prove the statement, make the substitution \( t = x - 1 \), set \( R(t) = P(t + 1) = a_{n}(t + 1)^{n} + a_{n - 1}(t + 1)^{n - 1} + \ldots + a_{1}(t + 1) + a_{0} \) and prove that \( R(t) \) is divisible by \( t^{2} \), i.e., that the last two coefficients of the polynomial \( R(t) \) are zero.

The constant term of the polynomial \( R \) is \( R(0) = a_{n} + a_{n - 1} + \ldots + a_{0} = 0 \).

Since in the polynomial \( (t + 1)^{k} \) the coefficient at \( t \) is \( k \), the coefficient at \( t \) of the polynomial \( R \) is \( n a_{n} + (n - 1) a_{n - 1} + \ldots + a_{1} \). From the conditions \( a_{n - k} = a_{k} \), it follows that the doubled coefficient at \( t \) is \( \left(n a_{n} + (n - 1) a_{n - 1} + \ldots + a_{1}\right) + \left(n a_{0} + (n - 1) a_{1} + \ldots + a_{n - 1}\right) = n\left(a_{n} + a_{n - 1} + \ldots + a_{0}\right) = 0 \).

Thus, the problem is solved.

Remark. The statement about the divisibility of \( P(x) \) by \( (x - 1)^{2} \) can be proven in several other ways. For example: 1) One can prove that when dividing \( P(x) \) by \( x - 1 \), we obtain a polynomial \( Q(x) = b_{n - 1} x^{n - 1} + b_{n - 2} x^{n - 2} + \ldots + b_{1} x + b_{0} \) such that \( b_{n - 1-k} = -b_{k} \) for all \( 0 \leqslant k \leqslant n - 1 \). From this follows \( Q(1) = 0 \), and then, by virtue of Bezout's theorem, \( Q(x) \) is divisible by \( (x - 1) \).

2) One can note that \( P(1) = 0 \) and \( P^{\prime}(1) = 0 \).

Let \( I \) be the center of the common incircle \( \omega \) of triangles \( A B C \) and \( D E F \). Since \( D \) is the midpoint of arc \( B C \), points \( A, I, D \) are collinear. The circle \( \omega \) is inscribed in angle \( \angle F D E \), so \( D I \) is the angle bisector of \( \angle F D E \), and point \( A \) is the midpoint of arc \( F E \).

Note that the quadrilateral \( F E Y X \) is cyclic. This follows from the equality of angles \( \angle F E D = \frac{1}{2}(\overparen{F B} + \overparen{B D}) = \frac{1}{2}(\overparen{F B} + \overparen{C D}) = \angle F X B \). Similarly, the quadrilateral \( B C T Z \) is cyclic.

If \( B C \parallel E F \), then the construction is symmetric with respect to the line \( A D \), and the statement of the problem is obvious. Otherwise, note the point \( S \) where lines \( F E \) and \( B C \) intersect. By equating the products of the segments of the secants for the circles \( \Omega, (B C T Z) \) and \( (F E Y X) \) (i.e., the power of point \( S \) with respect to these three circles), we have: \( S X \cdot S Y = S F \cdot S E = S B \cdot S C = S Z \cdot S T \). The proven equality \( S X \cdot S Y = S Z \cdot S T \) means that \( X, Y, Z, T \) lie on the same circle, which was to be proven.

Let \( r \) be the remainder of the smallest of the given odd numbers when divided by 2022, so \( r \) is some odd number from the set \( \{1,3,5, \ldots, 2021\} \). If \( r \leqslant 2017 \), then the other two remainders are \( -r + 2 \) and \( r + 4 \), so the sum of the remainders is \( r + (r + 2) + (r + 4) = 3(r + 2) \) - this number is composite, as it is divisible by 3 and greater than 3. Let's separately consider the cases \( r = 2019 \) and \( r = 2021 \). In the first case, the remainders of the given numbers are 2019, 2021, and 1. In the second case, the remainders of the given numbers are 2021, 1, and 3. In both cases, the sum of the remainders is divisible by 3 and greater than 3.

Remark 1. In all three cases, the sum of the three remainders is \( 3(r + 2) - 2022 k \), where \( k \in \{0,1,2\} \).

Remark 2. If in the problem statement 2022 is replaced by another number (not divisible by 3), the statement of the problem may become false. For example, the sum of the remainders of the numbers \( 19, 21, 23 \) when divided by 20 is \( 19 + 1 + 3 = 23 \) - a prime number.

Let \(\angle ABC = \alpha\), then by the condition \(\angle DAB = 2\alpha\). Extend segment \(AB\) beyond point \(A\) to point \(F\) such that \(AD = AF\). Then triangle \(AFD\) is isosceles, and its base angles are equal. Since \(\angle FAD = 180^{\circ} - 2\alpha\), it follows that \(\angle AFD = \angle ADF = \alpha\).

Since quadrilateral \(ABCD\) is cyclic, we have \(\angle ADC = 180^{\circ} - \angle ABC = 180^{\circ} - \alpha = 180^{\circ} - \angle ADF\). Therefore, points \(C, D,\) and \(F\) are collinear. Then \(\angle CFB = \alpha = \angle CBF\), so triangle \(FCB\) is isosceles. Thus, its bisector \(CE\) coincides with the median. Consequently, \(BE = EF = AD + AE\), which is what was required to prove.

Note. There is also a variation of the solution. Let \(\angle ABC = \alpha\), then by the condition \(\angle DAB = 2\alpha\). From the cyclic nature of quadrilateral \(ABCD\), we have \(\angle BCD = 180^{\circ} - \angle DAB = 180^{\circ} - 2\alpha\). Then \(\angle BCD + \angle ABC = 180^{\circ} - \alpha\), and the rays \(BA\) and \(CD\) intersect at some point \(F\), with \(\angle BFC = \alpha\). Since \(\angle CFB = \alpha = \angle CBF\), triangle \(FCB\) is isosceles. Thus, its bisector \(CE\) coincides with the median, so \(BE = EF = AF + AE\).

To complete the solution, it remains to show that \(AF = AD\). This follows from the cyclic nature of quadrilateral \(ABCD\), since \(\angle ADF = \angle CBF = \alpha\). Therefore, triangle \(AFD\) is isosceles with equal angles \(\angle AFD = \angle ADF = \alpha\).

Example. Take counter 50 and successively swap it 99 times with the next counter in the counterclockwise direction. This produces the required arrangement.

There are several ways to prove the estimate; below we give two of them.

First method. Suppose that for some \(k < 50\) the desired arrangement has been obtained.

At each moment, consider the arc colored that runs from counter 100 to counter 1 in the clockwise direction. Since counters 100 and 1 cannot be swapped in a single move, any particular counter \(m\) \((2 \leqslant m \leqslant 99)\) could enter the colored arc or leave the colored arc only by swapping with one of counters 1 or 100.

Since initially and in the final position counter \(m\) is not on the colored arc, it makes the same number of entries into and exits from the colored arc. For \(m \leqslant 50\), counter \(m\) cannot be swapped with counter 100, so it can enter or leave the colored arc only by swapping with counter 1. When \(m\) enters the arc, counter 1 shifts by 1 step clockwise; when \(m\) leaves, counter 1 shifts by 1 step counterclockwise. We carry out analogous reasoning for counters \(m \geqslant 51\), which cannot be swapped with counter 1.

Thus we conclude that counters 1 and 100 make the same total shift clockwise and counterclockwise, so they remain in their original positions. This is a contradiction.

Second method. We measure shifts of counters relative to their initial positions, counting a clockwise shift as positive and a counterclockwise shift as negative. Then, when we swap two counters, the shift of one increases by \(+1\), and that of the other decreases by \(-1\). Hence, after all operations the sum of all shifts is 0.

Argue by contradiction: assume that for \(k < 50\) each counter \(i\) is finally shifted by one position clockwise, i.e. its shift equals \(1 + 100 t_{i}\) (here \(t_{i}\) is an integer “number of full turns” clockwise; in particular, if \(t_{i} < 0\), then counter \(i\) has made \(\left|t_{i}\right|\) full turns counterclockwise). Then the total shift of all 100 counters is \(100\left(t_{1}+t_{2}+\ldots+t_{100}\right)+100\). Since this total must be 0, we get \(t_{1}+t_{2}+\ldots+t_{100}=-1\).

Since \(k < 50\), counters with numbers \(i\) and \(j\), where \(j \geqslant i+50\), could not be swapped, so at any moment their shifts differ by less than 100; therefore, their “numbers of turns” \(t_{i}\) and \(t_{j}\) are equal whenever \(j \geqslant i+50\). Hence \(t_{1}=t_{51}\), \(t_{2}=t_{52}, \ldots, t_{50}=t_{100}\). Thus the sum \(t_{1}+t_{2}+\ldots+t_{100}=2\left(t_{1}+t_{2}+\ldots+t_{50}\right)\) is even and therefore cannot be equal to \(-1\). Contradiction.

Remark 1. The last reasoning can be modified as follows. From the equalities we get \(t_{1}=t_{51}, t_{1}=t_{52}, \ldots, t_{1}=t_{100}, t_{2}=t_{100}, t_{3}=t_{100}, \ldots, t_{50}=t_{100}\); thus all \(t_{i}\) are equal to \(t_{1}\). Then \(t_{1}+t_{2}+\ldots+t_{100}=100 t_{1} \neq -1\).

Remark 2. One can also complete the contradiction argument in another way, replacing the last paragraph of the solution by the following reasoning. Color red those counters for which \(t_{i} \geqslant 0\), and blue those for which \(t_{i} < 0\). Clearly, at some move a blue counter and a red counter must be swapped, because the difference of their shifts is at least 100. Since \(k < 50\), the pair of counters 1 and 51 has the same color; similarly, the pairs 1 and 52, \ldots, 1 and 100, 2 and 100, 3 and 100, \ldots, 50 and 100 all have the same color. Thus all counters have the same color. But we know that \(t_{1}+t_{2}+\ldots+t_{100}=-1\), so among the numbers \(t_{1}, \ldots, t_{100}\) there are both nonnegative and negative ones, i.e. there are both red and blue counters — a contradiction.

From the problem statement, it follows that points \( A, A', B, B' \) lie in the same plane, so lines \( BA' \) and \( AB' \) intersect the edge \( CD \) at a single point \( X \). From the condition, these lines are the bisectors of angles \( CA'D \) and \( CB'D \) respectively. Hence, by the property of the bisector, \( CA' : A'D = CX : XD = CB' : B'D \), and since \( \angle CA'D = \angle CB'D = 120^{\circ} \), triangles \( CA'D \) and \( CB'D \) are similar. Since \( CD \) is the common side of these triangles, these triangles are congruent. In these congruent triangles, the corresponding altitudes from vertices \( A' \) and \( B' \) are equal. This is what needed to be proved.

It is stated that lines \( BA' \) and \( AB' \) intersect the edge \( CD \) at a single point; furthermore, the relation \( CA' : A'D = CB' : B'D \) or the similarity \( CA'D \sim CB'D \) is proven - 4 points.

In all the solutions below, we consider the friendship graph, where vertices are people in the company, and two people are connected by an edge if they are friends.

Consider 102 vertices, and construct the following graph on them. Connect one vertex \( x \) with three other vertices \( v_{1}, v_{2}, v_{3} \). The remaining 98 vertices are divided into pairs, and the vertices in each pair are connected. This results in a graph with \( 98 / 2 + 3 = 52 \) edges. When any vertex is removed, an odd number of edges is removed, so an odd number remains. Therefore, the company described in the problem can consist of 102 people.

It remains to show that there cannot be such a company with 103 people (and thus, there cannot be a company with more than 103 people). Below we provide several different ways to do this; in each method, we assume, for contradiction, that such a company exists.

First solution. There are a total of \( n = C_{103}^{2} = 51 \cdot 103 \) ways to remove two vertices from 103, leaving 101. Number these ways from 1 to \( n \). Let \( a_{i} \) be the number of edges on the remaining 101 vertices in the \( i \)-th way; by assumption, all numbers \( a_{i} \) are odd, and hence their sum \( S \) is also odd (since the number \( n \) is odd).

On the other hand, consider any edge \( uv \). This edge is counted in the number \( a_{i} \) exactly when the vertices \( u \) and \( v \) are not removed in the \( i \)-th way, that is, when some pair from the remaining 101 vertices is removed. This happens in \( k = C_{101}^{2} = 50 \cdot 101 \) ways. Thus, each edge is counted in \( S \) an even number \( k \) of times, so \( S \) must be even. Contradiction.

Second solution. Call a vertex even if its degree is even, and odd otherwise. Consider two cases.

Case 1. Suppose the total number of edges in the graph is odd. Then, when we remove any pair of vertices, we must remove an even number of edges (so that the remaining number of edges is still odd). On the other hand, if we remove vertices of degrees \(d_{1}\) and \(d_{2}\), then the number of removed edges is \(d_{1}+d_{2}\) if these vertices are not adjacent, and \(d_{1}+d_{2}-1\) if they are adjacent. It follows that vertices of the same parity are never adjacent, and vertices of different parity are always adjacent.

Hence, if the graph has \(k\) even vertices, then the total number of edges equals \(k(103-k)\), which is even. But we assumed that this number is odd — a contradiction.

Case 2. Suppose the total number of edges in the graph is even. Similarly, we obtain that vertices of the same parity are always adjacent, and vertices of different parity are not adjacent. Therefore, if the graph has \(k\) even vertices, then the number of missing edges is \(k(103-k)\), which is even. Thus the total number of edges is \[ \binom{103}{2} - k(103-k) = 103 \cdot 51 - k(103-k), \] which is odd. But we assumed that this number is even.

Remark 1. Of course, there are other examples of a group of 102 people that satisfies the condition.

Remark 2. There is also the following variation of the second solution. Consider any 102 vertices and the induced subgraph on these vertices, and let it have \(k\) edges. Removing from them any vertex (say, of degree \(d\)), we obtain 101 vertices with an odd number of edges \(k-d\). Thus the degree of any vertex in our subgraph has parity different from the parity of \(k\), so the degrees of all 102 vertices have the same parity.

Now consider the entire graph on 103 vertices. Call a vertex even if, after its removal, the remaining graph has all vertex degrees even, and odd otherwise. Then any two vertices of the same parity are adjacent to exactly the same subset of the remaining vertices, while any two vertices of different parity are adjacent to complementary subsets among the 101 remaining vertices. From this, as in the second solution, it is easy to deduce that the graph is either a complete bipartite graph or the disjoint union of two complete graphs. After that, one can proceed exactly as in the second solution.

Call a triple of natural numbers \((x, y, z)\), whose elements belong to \(S\), good if the equality \(x^2 + y^2 + z^2 = 2(xy + yz + zx)\) holds.

Thus we need to find the maximum possible number of good triples.

Let us determine when a triple is good. Rewrite (*) as a quadratic equation in \(x\): \(x^2 - 2x(y+z) + (y-z)^2 = 0\).

Solving it, we obtain \(x = (y+z) \pm 2\sqrt{yz} = (\sqrt{y} \pm \sqrt{z})^2\), whence \(\sqrt{x} = \pm\sqrt{y} \pm\sqrt{z}\). In other words, a triple is good if and only if one of the numbers \(\sqrt{x}, \sqrt{y}, \sqrt{z}\) equals the sum of the other two.

Let \(s_1 < s_2 < \ldots < s_{100}\) be all the elements of the set \(S\). Put \(t_i = \sqrt{s_i}\). Let us estimate the number of good triples \((x, y, z)\) in which \(z\) is the largest number, that is, \(\sqrt{z} = \sqrt{x} + \sqrt{y}\).

Note that for any \(1 \leqslant i < j \leqslant 100\) there is at most one such triple in which \(\sqrt{x} = t_i\) and \(\sqrt{z} = t_j\) (from these values we recover \(\sqrt{y} = \sqrt{z} - \sqrt{x}\)). Therefore the number we seek is at most the number of such pairs \((i, j)\), that is, \(\binom{100}{2}\).

Similarly, the numbers of good triples in which \(x\) and \(y\) are the largest elements are also at most \(\binom{100}{2}\). Hence the total number of good triples does not exceed \(3 \cdot \binom{100}{2}\).

This bound is attained if we set \(s_i = i^2\), that is, \(t_i = i\): indeed, then for any \(i < j\) there exists a good triple \((s_i, s_{j-i}, s_j)\).

Remark 1. One can show that, for the bound to be attained, it is necessary that the equalities \(t_i = i\,t_1\) hold for all \(i\). Since \(1 \leqslant t_i \leqslant 100\), the example given above is the only one.

Remark 2. One can show that the following fact holds: the equality \(\sqrt{z} = \sqrt{x} + \sqrt{y}\) for natural numbers \(x, y, z\) is possible only when the ratios \(x/z\) and \(y/z\) are squares of rational numbers, that is, when \(x = a^2 t\), \(y = b^2 t\) and \(z = c^2 t\) for some natural \(a, b, c\) and \(t\) (where \(a + b = c\)).

From the condition, it follows that \( x, y \geqslant 1 \), since the product of the digits of a natural number cannot be negative. Therefore, the numbers \( n \) and \( n + 1 \) do not contain zeros in their decimal representation. Then these numbers differ only by the last digit, and in the number \( n + 1 \) it is greater by one. Thus, \( y > x \).

If \( x - 1 > 0 \), then, reasoning similarly, we obtain that \( y - 1 < x - 1 \), which contradicts what was proven above. Therefore, \( x - 1 = 0 \). Then \( x = 1 \), and in the decimal representation of the number \( n \), all digits are 1. It follows that in the number \( n + 2 \), the last digit is two, and the other digits are ones, so \( y = 2 \). Hence, \( y - 1 = 1 \), and the number \( m \) consists only of ones. But then the number \( m + 1 \) does not contain zeros in its decimal representation. However, the product of its digits is zero, a contradiction.

Let \(O\) be the center of the circle circumscribed around the trapezoid \(ABCD\). Then \(\angle COD = 2 \angle CAD = \angle PAD + \angle ADP = \angle CPD\). Here we used the fact that the central angle is twice the inscribed angle, and that an exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles.

Therefore, point \(O\) lies on the circle \(\gamma\) circumscribed around triangle \(CPD\), and since \(OC = OD\), \(O\) is the midpoint of the arc \(CPD\). Then segment \(ON\) is the diameter of the circle \(\gamma\), and line \(ON\) is the perpendicular bisector of segment \(CD\). In particular, the midpoint of segment \(CD\), denote it by \(S\), lies on segment \(ON\). From the above, \(\angle OCN = 90^{\circ} = \angle CSN\). Thus, the circle circumscribed around triangle \(SCN\) is tangent to line \(OC\), so \(OS \cdot ON = OC^{2} = OK \cdot OL\). Let us denote point \(S'\), symmetric to point \(S\) with respect to point \(O\). Then \(OK \cdot OL = OS' \cdot ON\), so points \(S', K, L, N\) lie on the same circle.

Now notice that points \(M\) and \(S\) are symmetric with respect to line \(KL\). Thus, \(OS' = OS = OM\) and \(\angle LOS' = \angle KOS = \angle KOM\). Therefore, points \(M\) and \(S'\) are symmetric with respect to the perpendicular bisector of \(KL\). Consequently, points \(K, M, S'\), and \(L\) lie on the same circle. From the above, point \(N\) also lies on this circle, which is what was required.

First solution. Note that

\[ \frac{a^{3}}{a^{2} + b + c} = a-\frac{a(b + c)}{a^{2} + b + c} \geqslant a-\frac{a(b + c)}{2 a \sqrt{b + c}} = a-\frac{\sqrt{b + c}}{2} . \]

Here we estimated the denominator using the inequality of means: \[ a^{2} + b + c \geqslant 2 a \sqrt{b + c} \]

We sum the obtained inequality with three similar ones. Now it is sufficient to prove that

\[ a + b + c + d-\frac{\sqrt{a + b}}{2}-\frac{\sqrt{b + c}}{2}-\frac{\sqrt{c + d}}{2}-\frac{\sqrt{d + a}}{2} \geqslant 4 \]

Since \( a + b + c + d = 8 \), this is equivalent to the inequality

\[ \frac{\sqrt{a + b}}{2} + \frac{\sqrt{b + c}}{2} + \frac{\sqrt{c + d}}{2} + \frac{\sqrt{d + a}}{2} \leqslant 4 . \]

But from the inequality between the arithmetic mean and the quadratic mean, we obtain that

\[ \frac{\sqrt{a + b}}{2} + \frac{\sqrt{c + d}}{2} \leqslant \sqrt{\frac{(\sqrt{a + b})^{2} + (\sqrt{c + d})^{2}}{2}} = 2 \]

and, similarly,

\[ \frac{\sqrt{b + c}}{2} + \frac{\sqrt{d + a}}{2} \leqslant 2 . \]

Adding these two inequalities, we obtain the required result.

Second solution. By the Cauchy–Bunyakovsky–Schwarz inequality, we have

\(\displaystyle \frac{a^{3}}{a^{2}+b+c}+\frac{b^{3}}{b^{2}+c+d} +\frac{c^{3}}{c^{2}+d+a}+\frac{d^{3}}{d^{2}+a+b} \geqslant \frac{\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}}{a\left(a^{2}+b+c\right)+b\left(b^{2}+c+d\right)+c\left(c^{2}+d+a\right)+d\left(d^{2}+a+b\right)}. \)

Thus it is enough to prove that

\(\displaystyle \left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2} \geqslant 4\left(a^{3}+b^{3}+c^{3}+d^{3}+ab+bc+cd+da+2ac+2bd\right). \)

Note that

\(\displaystyle 32=\frac{(a+b+c+d)^{2}}{2} =ab+bc+cd+da+ac+bd+\frac{a^{2}+c^{2}}{2}+\frac{b^{2}+d^{2}}{2} \geqslant ab+bc+cd+da+2ac+2bd, \)

so it is enough to check that

\(\displaystyle \left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2} \geqslant 4\left(a^{3}+b^{3}+c^{3}+d^{3}+32\right). \) \(\quad (\star)\)

Make the substitution

\(\displaystyle a=2+x,\quad b=2+y,\quad c=2+z,\quad d=2+t. \)

Then

\(\displaystyle x+y+z+t = a+b+c+d-8 = 0, \)

\(\displaystyle a^{2}+b^{2}+c^{2}+d^{2} =4\cdot 2^{2}+2\cdot 2(x+y+z+t)+x^{2}+y^{2}+z^{2}+t^{2} =16+x^{2}+y^{2}+z^{2}+t^{2}, \)

\(\displaystyle a^{3}+b^{3}+c^{3}+d^{3} =4\cdot 2^{3}+3\cdot 2^{2}(x+y+z+t) +3\cdot 2\left(x^{2}+y^{2}+z^{2}+t^{2}\right) +x^{3}+y^{3}+z^{3}+t^{3}. \)

Inequality \((\star)\) takes the form

\(\displaystyle \left(x^{2}+y^{2}+z^{2}+t^{2}+16\right)^{2} \geqslant 4\left(x^{3}+y^{3}+z^{3}+t^{3} +6\left(x^{2}+y^{2}+z^{2}+t^{2}\right)+64\right). \)

After expanding and simplifying, it remains to prove that

\(\displaystyle \left(x^{2}+y^{2}+z^{2}+t^{2}\right)^{2} +8\left(x^{2}+y^{2}+z^{2}+t^{2}\right) \geqslant 4\left(x^{3}+y^{3}+z^{3}+t^{3}\right). \)

It remains to observe that

\(\displaystyle \left(x^{2}+y^{2}+z^{2}+t^{2}\right)^{2} +8\left(x^{2}+y^{2}+z^{2}+t^{2}\right) \geqslant x^{4}+y^{4}+z^{4}+t^{4} +4x^{2}+4y^{2}+4z^{2}+4t^{2} \geqslant 4\left(x^{3}+y^{3}+z^{3}+t^{3}\right). \)

The first inequality is obtained by expanding the brackets: after cancellation, only nonnegative terms remain on the left-hand side. The second is obtained by adding four mean-type inequalities of the form \(p^{4}+4p^{2} \geqslant 4p^{3}\).