6. C1 (UKR) On a $5 \times 5$ board, two players alternately mark numbers on empty cells. The first player always marks 1 's, the second 0 's. One number is marked per turn, until the board is filled. For each of the nine $3 \times 3$ squares the sum of the nine numbers on its cells is computed. Denote by A the maximum of these sums. How large can the first player make $A$, regardless of the responses of the second player?
6. Call the first and second player $M$ and $N$ respectively. $N$ can keep $A \leq 6$. Indeed, let 10 dominoes be placed as shown in the picture, and whenever $M$ marks a 1 in a cell of some domino, let $N$ mark 0 in the other cell of that domino if it is still empty. Since any $3 \times 3$ square contains at least three complete dominoes, there are at least three 0's inside. Hence $A \leq 6$.
We now show that $M$ can make $A=6$. Let him start by marking 1 in c3. By symmetry, we may assume that $N$ 's response is made in row 4 or 5 . Then $M$ marks 1 in $c 2$. If $N$ puts 0 in $c 1$, then $M$ can always mark two 1 's in $b \times\{1,2,3\}$ as well as three 1 's in $\{a, d\} \times\{1,2,3\}$. Thus either $\{a, b, c\} \times\{1,2,3\}$ or $\{b, c, d\} \times\{1,2,3\}$ will contain six 1's. However, if $N$ does not play his second move in $c 1$, then $M$ plays there, and thus he can easily achieve to have six 1's either in $\{a, b, c\} \times\{1,2,3\}$ or $\{c, d, e\} \times\{1,2,3\}$