Croatia Mathematical Competitions

Assume $n = 4k$. Then Ante can achieve that all numbers on the board are equal in the following way: he will increase by $1$ the numbers $1$ and $3$, $5$ and $7$, ..., $4k-3$ and $4k-1$. By doing that, he gets that the numbers on the board are all even numbers smaller than or equal to $n$, and each is written twice. Finally, he increases $2$ and $2$ by $n-2$, $4$ and $4$ by $n-4$, ..., $n-2$ and $n-2$ by $2$, and he gets that all numbers on the board are equal to $n$.

Assume $n = 2k+1$. Then Ante can achieve that all numbers on the board are equal in the following way: he will increase by $1$ the numbers $1$ and $n$, $3$ and $n$, ..., $n-2$ and $n$. By doing that, he gets that the numbers on the board are all even numbers smaller than or equal to $n$, each written twice, and the number $\frac{3n-1}{2}$. Finally, he increases $2$ and $2$ by $\frac{3n-5}{2}$, $4$ and $4$ by $\frac{3n-9}{2}$, ..., $n-1$ and $n-1$ by $\frac{n+1}{2}$. Now all numbers on the board are equal to $\frac{3n-1}{2}$.

Assume $n = 4k + 2$. Then Ante cannot achieve that all numbers are equal. The sum of all numbers on the board is initially odd, because
$$1 + 2 + \dots + n = \frac{n(n+1)}{2} = (2k+1)(4k+3).$$
Since there is an even number of numbers on the board, if they were equal their sum would be an even number. On the other hand, in each step the sum of numbers on the board is increased by an even number, so the sum will never be even.

Therefore, Ante can achieve that all numbers are equal if and only if $n$ is not of the form $4k + 2$, $k \in \mathbb{N}$.