2. List the natural numbers $1, 2, \cdots, k^{2}$ in a square table (as shown in Table 1), then select any one number from the table, and subsequently remove the row and column containing that number. Then, perform the same operation on the remaining $(k-1)^{2}$ numbers in the square table, and continue this process $k$ times. The sum of the $k$ selected numbers is
1
\begin{tabular}{|c|c|c|c|}
\hline 1 & 2 & $\cdots$ & $k$ \\
\hline$k+1$ & $k+2$ & $\cdots$ & $2 k$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline$(k-1) k+1$ & $(k-1) k+2$ & $\cdots$ & $k^{2}$ \\
\hline
\end{tabular}
2. $\frac{1}{2} k\left(k^{2}+1\right)$.

Divide Table 1 into the following two number tables:
It is easy to see that each number in Table 1 equals the sum of the two numbers in the same position in the divided Table 2 and Table 3. Therefore, the sum of the $k$ selected numbers, which are neither in the same row nor in the same column, is exactly
$$
\begin{array}{l}
S=(1+2+\cdots+k)+[0+k+\cdots+(k-1) k] \\
=\frac{1}{2} k(k+1)+\frac{1}{2} k^{2}(k-1)=\frac{1}{2} k\left(k^{2}+1\right) .
\end{array}
$$