Ukrainian National Mathematical Olympiad

We will show that there exists a strategy such that Arsenii always has a turn after Andrii's turn. We will split all the cells into friendly pairs. Two $1 \times 1$ cells make a friendly pair, if they are in the same column and there are exactly two $1 \times 1$ cells in-between. Then Arsenii colors the cell that makes a friendly pair with a cell that Andrii colored during his last turn.

We want to show that Arsenii always has such a turn. If Andrii colored some cell $P$, then a cell $V$, that makes a friendly pair with $P$ is white. Moreover, suppose the cell $V$ cannot be colored in black because there already is another black cell $Y$ in some $3 \times 3$ square that is located either in the very top or the very bottom of the table. Then there also exists another similar $3 \times 3$ square either in the very top or the very bottom that contains cells $P$ and $Y$, both of which are black, where $Y$ is a friendly pair of $V$ (Fig. 3). This completes the proof.