[Solution] Let 1 be placed in the cell at the $i$-th row and $j$-th column. When the $i$-th column already has $n-1$ numbers, the next number will fill the $i$-th row. Therefore, to ensure the process of filling numbers can proceed to the end, $n^{2}$ must be in the $i$-th column.

Except for the $i$-th row where the number 1 is located, the sum of the numbers in each of the other rows is $n$ more than the sum of the numbers in the corresponding column. The sum of the numbers in the $n-1$ rows is $n(n-1)$ more than the sum of the numbers in the corresponding $n-1$ columns. Since the sum of the numbers in the $n$ rows is equal to the sum of the numbers in the $n$ columns, the sum of the numbers in the column where $n^{2}$ is located is $n(n-1)$ more than the sum of the numbers in the row where 1 is located.
$3 \cdot 13$ Fill all integers from 1 to $n^{2}$ into an $n \times n$ square grid according to the following rules: 1 can be placed in any cell, the row of 2 should have the same column index as the column of 1, the row of 3 should have the same column index as the column of 2, and so on. What is the difference between the sum of all numbers in the row where 1 is located and the sum of all numbers in the column where $n^{2}$ is located?