The columns of an $ n\times n$ board are labeled $ 1$ to $ n$. The numbers $ 1,2,...,n$ are arranged in the board so that the numbers in each row and column are pairwise different. We call a cell "good" if the number in it is greater than the label of its column. For which $ n$ is there an arrangement in which each row contains equally many good cells?
1. **Claim**: \( n \) must be odd for the arrangement to exist.
   
2. **Proof for \( n \) being even**:
   - Consider an \( n \times n \) board where \( n \) is even.
   - In the \( i \)-th column, there are \( i-1 \) good cells because a cell is good if the number in it is greater than the column label.
   - Summing up the good cells for all columns, we get:
     \[
     \sum_{i=1}^{n} (i-1) = 0 + 1 + 2 + \cdots + (n-1) = \frac{n(n-1)}{2}
     \]
   - For each row to contain an equal number of good cells, the total number of good cells \( \frac{n(n-1)}{2} \) must be divisible by \( n \).
   - Since \( n \) is even, \( \frac{n(n-1)}{2} \) is not divisible by \( n \) because \( n-1 \) is odd and \( n \) is even, making \( \frac{n-1}{2} \) not an integer.
   - Therefore, \( n \) cannot be even.

3. **Constructing the arrangement for odd \( n \)**:
   - Label the rows from top to bottom as \( 1, 2, \ldots, n \).
   - Consider the arrangement where in the \( i \)-th column, each natural number \( k \in \{1, 2, \ldots, n\} \) is placed in row \( (k-2i) \mod n \).
   - For example, for \( n = 5 \), the arrangement is:
     \[
     \begin{array}{ccccc}
     1 & 3 & 5 & 2 & 4 \\
     2 & 4 & 1 & 3 & 5 \\
     3 & 5 & 2 & 4 & 1 \\
     4 & 1 & 3 & 5 & 2 \\
     5 & 2 & 4 & 1 & 3 \\
     \end{array}
     \]
   - In this arrangement, each row contains \( \frac{n-1}{2} \) good cells.
   - Each number \( 1, 2, \ldots, n \) appears exactly once in each row and each column.

4. **Conclusion**:
   - There exists an arrangement where each row contains equally many good cells if and only if \( n \) is odd.

\(\blacksquare\)

The final answer is \( \boxed{ n } \) is odd.