4. $A$ and $B$ are playing a game on a $25 \times 25$ grid. At the beginning, $A$ can mark some cells. After marking is completed, they start taking turns to place gold coins on the grid, with $B$ going first. The rules for placing gold coins are as follows: (1) gold coins cannot be placed in marked cells; (2) once a gold coin is placed, no more gold coins can be placed in the same row or column; (3) if no more gold coins can be placed on the grid, the game ends. The player who places the last gold coin on the grid wins. How many cells does $A$ need to mark at the beginning of the game to ensure a win? (Assume both $A$ and $B$ use the best strategy)
Four, Analysis: Generalize the problem, and prove the case of playing the game on a $(2 n+1) \times(2 n+1)$ grid.

On the one hand, if $A$ marks $2 n+1$ squares, there is a winning strategy: $A$ marks all the squares in the first row. Then, in the remaining $2 n \times(2 n+1)$ squares, each operation will result in one row where no more coins can be placed. Since the remaining number of rows is even, the second player must win, i.e., $A$ must win.

On the other hand, it can be proven that when $A$ marks fewer than $2 n+1$ squares, $B$ has a winning strategy. The following proof uses mathematical induction:
(1) When $n=0$, $A$ cannot mark any squares, and it is easy to see that the first player wins;
(2) Assume that when $n=k$, i.e., in a $(2 k+1) \times(2 k+1)$ grid, if $A$ marks fewer than $2 k+1$ squares, then $B$ has a winning strategy. When $n=k+1$, at most $2 k+2$ squares are marked.
(1) If two marked squares are in the same row or column, $B$ chooses to place a coin in that row.
(2) If no two marked squares are in the same row or column, assume $A$ marks the square at the $i$-th row and $j$-th column, and the square at the $s$-th row and $t$-th column. In this case, $B$ places a coin at the $i$-th row and $s$-th column.

In this case, no matter how $A$ operates, in the remaining $(2 k+1) \times(2 k+1)$ squares, there are at most $2 k$ marked squares. By the induction hypothesis, $B$ has a winning strategy.

In summary, at least $2 n+1$ squares must be marked. Returning to the original problem, from the above process, it is known that $A$ must mark at least 25 squares.