For the sum of two numbers to be divisible by 6, their remainders modulo 6 must complement each other to 6. Possible pairs of remainders are:

• 0 and 0,
• 1 and 5,
• 2 and 4,
• 3 and 3.

Thus, all numbers divisible by 6 (remainder 0) can only be paired with each other. Therefore, the number of such numbers must be even. However, in the range from 1 to 800, there are exactly ⌊800/6⌋ = 133 numbers divisible by 6 (these are the numbers 6, 12, ..., 798). Since 133 is odd, it is not possible to pair all 800 numbers with the required property.

Let the numbers around the circle be \(a_1, a_2, \, \ldots, \, a_{17}\). Then the new numbers will be \(a_2 - a_3, a_3 - a_4, \, \ldots, \, a_{17} - a_1, a_1 - a_2\), and their sum is 0.

If the sum of 17 numbers is even, then there is at least one even number among them (the sum of an odd number of odd numbers is odd). Thus, the product of these numbers is even.

Answer: No.

The number A = M – N = 2N is even. However, according to the condition, the number A is composed of odd digits and a two. Therefore, A ends with 2. Hence, the number N, which is half of A, ends with either 1 or 6.

If N ends with 1, doubling it does not cause a carry from the last to the second-to-last digit. Therefore, the second-to-last digit of A = 2N would be even, but it should be odd. Contradiction.

We record each of our differences with a positive sign if in the corresponding pair of numbers the larger number precedes the smaller one clockwise, and with a negative sign otherwise. We have 11 differences between a number and the next one clockwise; thus, the sum of all these numbers is zero, which means it is even. But this is impossible, since there are exactly seven odd numbers among them.

Consider the leftmost even number \(a_i\) and the rightmost even number \(a_k\). Notice that all sums with indices from \(i\) to \(k - 1\) are even, and only they are (in sums with smaller indices, the first term is odd and the second is even; in sums with larger indices, the opposite is true).

Thus, \(k - i = 32\). The number of even numbers will be maximized if all numbers between \(a_i\) and \(a_k\) are also even. In this case, the number of even numbers will be 33.

Let there be \( a \) even numbers and \( b = 14 - a \) odd numbers among our 14 numbers. An odd number on the board can only appear as the sum of an even and an odd number, so there will be \( ab \) such numbers (each will be written twice). But \( 4ab \leq (a + b)^2 = 4 \cdot 49 \). Therefore, there will be at most 49 different odd numbers on the board; however, to satisfy the condition, there must be at least 50.

Assume initially the sum included the fraction \( \frac{a}{2} \). We will show that in the original sum, there exists a fraction \( \frac{b}{c} \) with an odd denominator \( c \) such that the numbers \( a \) and \( b \) have different parities. Indeed, there are exactly 50 fractions with odd denominators, and the number \( a \) is not the numerator of any of them. Therefore, among the numerators of such fractions, at most 49 have the same parity as \( a \).

Now, swap the numerators \( a \) and \( b \). Do this in two steps: first, swap the numerator of the fraction with denominator 2 (the sum changes by an odd number \( a - b \) of halves and thus becomes an integer), and then swap the numerator of the fraction with denominator \( c \) (the sum changes by a fraction with an odd denominator, thus becoming a fraction with an odd denominator).

Let \( n = 2^t \cdot m \), where \( t \geq 0 \), and \( m \) is odd. We will prove that \( f(n) \) is even in exactly two cases: either \( n \) is odd and \( m \equiv 1 \pmod{4} \), or \( n \) is even and \( m \equiv 3 \pmod{4} \).

First method. Consider an arbitrary fraction \( \frac{k}{n} \). If \( k \) is divisible by \( 2^{t+1} \), then the numerator of this fraction after reduction will be even; otherwise, it will be odd. Among the numbers \( 1, 2, \ldots, n - 1 \), there are exactly \( \frac{m-1}{2} \) numbers divisible by \( 2^{t+1} \). Thus, in the sum \( f(n) \), there are exactly \( n - 1 - \frac{m-1}{2} \) odd terms. Therefore, \( f(n) \) is even if and only if the numbers \( n - 1 \) and \( \frac{m-1}{2} \) have the same parity, as required.

Second method. Suppose \( n \) is odd (i.e., \( t = 0 \)). Then the numerator of the fraction \( \frac{k}{n} \) does not change parity after reduction. Thus, the number of odd numerators will be \( \frac{n-1}{2} \), and \( f(n) \) is even if and only if \( \frac{n-1}{2} \) is even, which occurs when \( m \equiv 1 \pmod{4} \).

Suppose \( n \) is even (\( t > 0 \)). Among the fractions with denominator \( n \), there is a fraction equal to \( \frac{1}{2} \) contributing 1 to \( f(n) \). All other fractions are paired as \( \left(\frac{a}{n}, \frac{n-a}{n}\right) \). Since the sum of fractions in a pair is 1, after reduction, they become pairs of irreducible fractions with the same denominator \( \left(\frac{b}{d}, \frac{d-b}{d}\right) \). The contribution of such a pair to \( f(n) \) is \( d \), and if \( d \) is even, it does not affect the parity of \( f(n) \).

Thus, the parity of \( f(n) \) is opposite to the parity of a similar sum for fractions, i.e., the parity of \( f(m) \) (for \( m = 1 \), it is opposite to the parity of \( f(1) = 0 \)). This gives the required result.

Finally, note that the numbers \( n \) and \( 2015n \) have the same parity; moreover, for odd \( m \), the numbers \( m \) and \( 2015m \) leave different remainders when divided by 4. Therefore, the numbers \( f(n) \) and \( f(2015n) \) always have different parity.

Assume there is an even number of negative numbers among the given numbers. Then there is a positive number \(a\), and the product of all numbers except \(a\) is positive. This contradicts the condition. Therefore, there is an odd number of negative numbers among the given numbers.

Let \(x_1, x_2, \ldots, x_k\) and \(y_1, y_2, \ldots, y_m\) be two groups into which the given numbers are divided \((k + m = 2011)\). Exactly one of the two products \(x_1x_2\ldots x_k\) and \(y_1y_2\ldots y_m\) (specifically, the one with an odd number of negative factors) is negative; let \(x_1x_2\ldots x_k < 0\), \(y_1y_2\ldots y_m > 0\). Among the numbers \(x_1, x_2, \ldots, x_k\), there will be a negative one, say \(x_1 < 0\). Then \(x_2\ldots x_k > 0\), and thus \(x_2\ldots x_k \geq 1\) (since the numbers are integers). Therefore,

\(x_1x_2\ldots x_k + y_1y_2\ldots y_m \leq x_1 + y_1y_2\ldots y_mx_2\ldots x_k < 0.\)