## Solution.

Let $n=4k$. Then Ante can achieve that all numbers on the board are equal in the following way: he increases by 1 the numbers 1 and 3, 5 and 7, ..., $4k-3$ and $4k-1$. This way, on the board, he gets all even numbers less than or equal to $n$ written twice. In the end, he increases 2 and 2 by $n-2$, 4 and 4 by $n-4$, ..., $n-2$ and $n-2$ by 2, and thus all numbers on the board are equal to $n$.

Let $n=2k+1$. Then Ante can achieve that all numbers on the board are equal in the following way: he increases by 1 the numbers 1 and $n$, 3 and $n$, ..., $n-2$ and $n$. This way, on the board, he gets all even numbers less than or equal to $n$ written twice and the number $\frac{3n-1}{2}$. In the end, he increases 2 and 2 by $\frac{3n-5}{2}$, 4 and 4 by $\frac{3n-9}{2}$, ..., $n-1$ and $n-1$ by $\frac{n+1}{2}$.

Let $n=4k+2$. Then Ante cannot achieve that all numbers are equal. The sum of all numbers on the board is initially odd, since

$$
1+2+\ldots+n=\frac{n(n+1)}{2}=(2k+1)(4k+3)
$$

During each step, the sum of the numbers on the board increases by an even number, so it will never be even as it would need to be if all numbers on the board (of which there is an even number) were equal.

Ante can achieve that all numbers on the board are equal if and only if $n$ is not of the form $4k+2$, $k \in \mathbb{N}$.
## Task A-1.4.

On the board, there are the first $n$ natural numbers $(n \geqslant 3)$. Ante repeats the following procedure: first, he arbitrarily selects two numbers on the board, and then increases them by the same arbitrary amount.

Determine all natural numbers $n$ for which Ante, by repeating this procedure, can achieve that all numbers on the board are equal.