Let $A$ and $B$ be the players. Suppose $A$ erases the number $2025$ on his first turn. Then, he pairs the following numbers:$$(1,2024), \quad (2,2023), \quad \ldots, \quad (1012, 1013).$$Every time $B$ erases the number $n$, $A$ will erase the number $2025-n$. Therefore, the last two numbers will belong to the same pair. Since $2025=45^2$ is a perfect square, $A$ has the winning strategy.

Mykhailo should erase $2025$ first, then whenever Oleksii chooses a number, say $o$, Mykhailo chooses $2025 - o$. The sum of the final two numbers will be $2025 = 45^2$, which is a perfect square, so Mykhailo can guarantee a win.