In general, for an $n \times n$ table, we claim the answer is $n$. First, we describe Gholam's strategy to achieve this score. Whenever Sahand presses a row button, Gholam in his next turn presses a column button. By the time Sahand presses a column button, Gholam presses a row button.

Suppose that Sahand picks $k$ rows and $n-k$ columns. Then from a column Gholam picks corresponding to Sahand's $i$-th row choice, at least $n-k+i$ cells remain red, since only rows Sahand picks after this can turn a cell in this column black. On the other hand, if Gholam picks a row which is the $j$-th row chosen, we get at least $j$ new red cells because $n-j$ columns remain, any of which could be black or already counted. Therefore, in total, we at least yield $(1+2+\cdots+n-k)+(n-k+1+n-k+2+\cdots+n) = \frac{n^2+n}{2}$ red cells, which is at least $n$ more than the black cells.

To show that Gholam cannot performs better, we must show Sahand has a strategy that ensures his score is at least $-n$. Suppose Sahand makes the first row black. From then on, whenever Gholam chooses a row, Sahand chooses a column, and vice-versa. Analogously, if Gholam chooses $k$ columns and his last choice is a column, at least $(1+2+\cdots+n-k)+(n-k+1+n-k+2+\cdots+n-1)$ cells become black. If Gholam's last choice is a row, at least $(1+2+\cdots+(n-k-1))+(n-k+1+n-k+2+\cdots+n-1+n)$ cells become black. Both these sums are at least $\frac{n^2-n}{2}$.

Other ways exist to prove this part, for example, by induction and analysis after Gholam's first move.