, 2022

The largest $x$ is $\frac{1}{50 \times 51}$. More generally, for an $n \times n$ chessboard, let $a = \lfloor \frac{n+1}{2} \rfloor$, $b = \lfloor \frac{n+2}{2} \rfloor$, then the largest $x$ is $\frac{1}{ab}$.

First, we construct a chessboard achieving this $x$. Let us divide the chessboard into four regions:
- The $a \times b$ cells in the upper-left corner are all filled with $\frac{1}{ab}$;
- The $(n-a) \times b$ cells in the upper-right corner are all filled with $\frac{1}{b}$;
- The $a \times (n-b)$ cells in the lower-left corner are all filled with $\frac{1}{a}$;
- The $(n-a) \times (n-b)$ cells in the lower-right corner are all filled with 0.

Direct verification shows that this chessboard is balanced (note that $a+b=n+1$). Moreover, according to the condition required by the problem of different rows and columns, if we select $n$ cells without selecting from the $(n-a) \times (n-b)$ cells in the lower-right corner, then we must necessarily need to select from the $a \times b$ cells in the upper-left corner, so $x$ must be at least $\frac{1}{ab}$.

Next we prove that $x = \frac{1}{ab}$ is already sufficient. Given a balanced chessboard, let us color black all cells with value $\ge \frac{1}{ab}$, and color the remaining cells white. We wish to prove: for any $p$ rows, the black cells on these $p$ rows will together appear in at least $p$ different columns. If this property holds, then by using the black cells to match the corresponding rows and columns, by Hall's theorem, there exists a perfect matching between rows and columns, that is, we can find $n$ black cells in different rows and columns, which completes the proof.

To prove this property, for any $p$ rows, suppose the black cells on these $p$ rows appear in $q$ different columns. Consider the cells that are in these $p$ rows but not in these $q$ columns, then they must all be white cells, so the sum of the numbers in these cells is $< \frac{1}{ab}p(n-q)$. Also, since the sum of numbers in each row is 1, the sum of the numbers in the cells that are in these $p$ rows and in these $q$ columns is $> p - \frac{1}{ab}p(n-q)$. But note that the sum of the numbers in these $q$ columns is $q$, and this number must not be less than the sum of the numbers in the cells that are in these $p$ rows and in these $q$ columns, so

$$q > p - \frac{1}{ab}p(n-q) \Rightarrow q > \frac{abp - pn}{ab - p}.$$

But it is also easy to see that
$$ab \ge p(n+1-p) \Rightarrow \frac{abp - pn}{ab - p} \ge p - 1,$$

so we have $q > p - 1 \Rightarrow q \ge p$. This proves the property to be proved, and thus proves the original proposition.