The answer is $51$. If a cell can be a corner of an $n\times n$ square, then one can find the number in that cell. Since for $n=51$, every cell can be a corner of an $n\times n$ square, Wolf can find the number in all cells. For $n\geq52$, for all $n\times n$ squares and $(n-1)\times1, 1\times(n-1)$ rectangles, the sum of which is the same on the two $100\times100$ tables with the following different $2\times2$ squares at the center. $\begin{array}{|c|c|}\hline a&b\\\hline c&d\\\hline\end{array}$ and $\begin{array}{|c|c|}\hline a+e&b-e\\\hline c-e&d+e\\\hline\end{array}$ Therefore, Wolf can't guarantee the numbers in the $4$ cell at the center of the board for $n\geq52$.