A $100 \times100$ chessboard has a non-negative real number in each of its cells. A chessboard is [b]balanced[/b] if and only if the numbers sum up to one for each column of cells as well as each row of cells. Find the largest positive real number $x$ so that, for any balanced chessboard, we can find $100$ cells of it so that these cells all have number
greater or equal to $x$, and no two of these cells are on the same column or row.

[i]Proposed by CSJL.[/i]
1. **Generalize the Problem**: Replace \(100\) with \(n\). We need to find the largest positive real number \(x\) such that for any balanced \(n \times n\) chessboard, we can find \(n\) cells such that these cells all have numbers greater than or equal to \(x\), and no two of these cells are in the same row or column.

2. **Formulate the Answer**: The answer is \(\frac{1}{\left(\frac{n+1}{2}\right)^2}\) if \(n\) is odd and \(\frac{1}{\frac{n}{2} \cdot \left(\frac{n}{2}+1\right)}\) if \(n\) is even.

3. **Construct the Chessboard**:
   - Let \(x = \lceil \frac{n+1}{2} \rceil\) and \(y = \lfloor \frac{n+1}{2} \rfloor\). Then \(x + y = n + 1\).
   - Define \(a_{i,j}\) as the cell on the \(i\)-th row and \(j\)-th column, where \(a_{1,1}\) denotes the top left corner.
   - Assign values to the cells as follows:
     \[
     a_{i,j} = 
     \begin{cases} 
     0 & \text{if } i < y \text{ and } j < x \\
     \frac{1}{x} & \text{if } i \ge y \text{ and } j < x \\
     \frac{1}{y} & \text{if } i < y \text{ and } j \ge x \\
     \frac{1}{xy} & \text{if } i \ge y \text{ and } j \ge x 
     \end{cases}
     \]

4. **Proof of Optimality**:
   - Consider a bipartite graph \((R, C)\) where \(R\) is the set of rows and \(C\) is the set of columns. Row \(i\) and column \(j\) are adjacent with an edge if \(a_{i,j} \ge \frac{1}{xy}\).
   - By Hall's Marriage Theorem, it suffices to show that for every \(X \subset R\) of rows, \(|N(X)| \ge |X|\).

5. **Contradiction Argument**:
   - Assume for contradiction that there exists a set of \(m\) rows (call them \(r_1, \ldots, r_m\)) such that \(|N(X)| \le m - 1\).
   - This implies there are at least \(n + 1 - m\) columns (call them \(c_1, \ldots, c_{n+1-m}\)) such that \(a_{j,k} < \frac{1}{xy}\) for every \(1 \le j \le m\) and \(1 \le k \le n + 1 - m\).
   - Let \(S\) be the set of squares in the intersection of these \(n + 1 - m\) columns and \(m\) rows. The sum of values of squares in \(S\) is at most \((n + 1 - m)m \frac{1}{xy} < 1\), and equality cannot hold.
   - Let \(T\) be the set of squares in the intersection of these \(m\) rows but the other \(m - 1\) columns. Since \(S\) and \(T\) form \(m\) rows, the sum of values of squares in \(S\) plus the sum of values of squares in \(T\) is exactly \(m\). It follows that the sum of numbers in squares \(S\) is greater than \(m - 1\). However, \(S\) is a subset of \(m - 1\) columns, which is a contradiction.

\(\blacksquare\)

The final answer is \( \boxed{ \frac{1}{\left(\frac{n+1}{2}\right)^2} } \) if \(n\) is odd and \(\frac{1}{\frac{n}{2} \cdot \left(\frac{n}{2}+1\right)}\) if \(n\) is even.