Denote $A_1,A_2, \cdots A_n$ be the row. We consider a step that send any number in row $A_k \; (1 \le k \le n)$ to a number in row $A_1$ by subtracting $n(k-1)$, we call this a subtracting number. Let $(x_1,x_2, \cdots , x_n)$ be the numbers that Peter picks. Since each number in the different row, this means each number needs a different subtracting numbers in order to do the step. Therefore, the total amount of subtracting number for Peter's numbers is always $n+2n+ \cdots + (n-1)n= \frac{(n-1)n^2}{2}$. Since Peter's number are in the different columns so the step will turn Peter's numbers to all numbers in $A_1$ after the step, which is $1,2, \cdots, n$. We have $S=\sum_{i=1}^n x_i = \frac{(n-1)n^2}{2}+ \frac{n(n+1)}{2}$. This shows that Peter always gets the same number for $S$, no matter how he chooses his $n$ numbers.