Belarus — Final Round

Answer: $n = 2^m$, where $m \in \mathbb{N}$, $m > 1$.

The goal of Alice is to obtain exactly $n$ sweets in any box. The number $a-n$ is said to be the distance if $a$ is a number of the sweets in the box. Note that the sum of the distances is equal to zero at any moment. Now we see that Alice's goal is to make all distances to be zeros. If there are two boxes with the distances $d_1$ and $d_2$ of the same parity, then Alice can decrease the absolute values of the distances for these boxes if and only if $d_1 \neq d_2$. The necessity of this inequality is obvious. To prove the sufficiency we suppose that $|d_1| \ge |d_2|$:

1) if $|d_1| > |d_2|$, then new distances of these boxes satisfy the inequality $|(d_1 + d_2)/2| \le (|d_1| + |d_2|)/2 < (|d_1| + |d_1|)/2 = |d_1|$;

2) if $d_1 = -d_2$, then new distances of these boxes are equal to zero.

Therefore, Alice cannot continue decreasing the absolute values of the distances if and only if at some moment either all distances are equal to zeros (and the goal is attained) or $k$ distances are even and equal to some $\alpha > 0$, and all other distances are odd and equal to some $\beta < 0$.

Let $n = 2^m$ ($m > 1$). The sum of all distances is equal to zero at any moment. So, if $k$ distances are even and equal to $\alpha$, and all other $n-k$ distances are odd and equal to $\beta$, then $\alpha k + \beta(n-k) = 0$ which yields $k(\alpha - \beta) = 2^m\beta$. But the last equality is impossible since $\alpha - \beta$ is odd and $k < 2^m$. Therefore, if $n = 2^m$ ($m > 1$), then Alice can equalize the numbers of the sweets in all boxes.

Now let $n \neq 2^m$. Then $n = (2p + 1) \cdot 2^m$ for some positive integer $p$ and nonnegative integer $m$. Suppose that there are $k = 2^m$ boxes with the distance $\alpha = 2p$ and all other $n-k = 2p2^m$ boxes with the distance $\beta = -1$. Note that this distribution of the sweets is possible since the equality $\alpha k + \beta(n-k) = 0$ holds. Since the numbers $\alpha$ and $\beta$ have the different parities, Alice cannot equalize the numbers of the sweets in all boxes.