Chuviiin K.

Given an $n \times n$ table, the columns of which are numbered from 1 to $n$. The cells of the table are filled with numbers $1, \ldots, n$ such that all numbers in each row and each column are distinct. We will call a cell good if the number in it is greater than the number of the column in which it is located. For which $n$ does there exist an arrangement in which all rows have the same number of good cells?
Let's find the total number of good cells. In the first column, there are $n-1$ (all except the cell with the number 1), in the second column - $n-2$ (all except the cells with the numbers 1 and 2), and so on, in the last column there are no such cells. Therefore, the total number is $(n-1)+(n-2)+\ldots+1=1 / 2 n(n-1)$. Hence, in each row, there should be ${ }^{n-1 / 2}$. Therefore, $n$ must be odd.

Let's provide an example of placement for $n=2 k+1$.

In each row, there is a cyclic permutation of the sequence $2 k+1,2 k, \ldots, 1$. In this case, as can be easily seen, in the $2 i$-th row, the first $i-1$ cells and the cells from the $2 i$-th to the $(k+i)$-th will be good, and in the $(2 i-1)$-th row - the first $i-1$ cells and the cells from the $(2 i-1)$-th to the $(k+i-1)$-th. Thus, in each row, there are exactly $k$ good cells.

## Answer

For odd $n$.