【Example 7】Arrange the natural numbers $1,2, \cdots, k^{2}$ into a square array:
\begin{tabular}{llllc}
1 & 2 & 3 & $\cdots$ & $k$ \\
$k+1$ & $k+2$ & $k+3$ & $\cdots$ & $2 k$ \\
$\vdots$ & $\vdots$ & $\vdots$ & & $\vdots$ \\
$(k-1) k+1$ & $(k-1) k+2$ & $(k-1) k+3$ & $\cdots$ & $k^{2}$
\end{tabular}

Select any number from the table, then delete all the numbers in the row and column containing this number. Next, from the remaining square array of $(k-1)^{2}$ numbers, select any number and delete all the numbers in the row and column containing this number. Continue this process until you have selected $k$ different numbers from the table, forming a $k$-element subset of $\left\{1,2, \cdots, k^{2}\right\}$. An interesting question is: how many different $k$-element subsets of $\left\{1,2, \cdots, k^{2}\right\}$ can be obtained using this method? What are the sums of the $k$ numbers in these subsets?
Since once a number is obtained, all the numbers in the row and column where this number originally was are deleted, it is impossible for any two of the $k$ numbers obtained to belong to the same row or the same column. In other words, these $k$ numbers are all taken from different rows and different columns. Conversely, any $k$ numbers that belong to different rows and different columns can be obtained in such a way. As long as the number belonging to the first row is taken first, then the first row and the column it is in are deleted, and the number belonging to the second row is taken, and so on, until all $k$ numbers are taken. 

Therefore, we only need to find out how many different $k$-tuples of numbers, which belong to different rows and different columns, there are in the square array. For this purpose, we first arrange these $k$ numbers in a column according to the natural order of their rows, and denote the column numbers they belong to as $j_{1}, j_{2}, \cdots, j_{k}$, then these column numbers are exactly a permutation of $1,2, \cdots, k$, so there are $A_{k}^{k}=k!$ such $k$-tuples.

Next, we calculate the sum of the $k$ numbers in these subsets. We know that the number in the square array located at the $i$-th row and $j$-th column is $(i-1) k+j$. However, since each subset of $k$ numbers includes one number from each row and each column, the sum of the $k$ numbers in each subset is
$$
\begin{array}{l}
{[0+k+2 k+\cdots+(k-1) k]+[1+2+3+\cdots+k]} \\
\quad=\frac{1}{2} k\left(k^{2}+1\right)
\end{array}
$$