Example 1 Find the sum of the $n^{2}$ numbers:
$$
\begin{array}{l}
1,2,3, \cdots, n \\
2,3,4, \cdots, n+1 \\
3,4,5, \cdots, n+2 \\
\cdots \cdots \\
n, n+1, n+2, \cdots, 2 n-1
\end{array}
$$
Analyze: In this matrix, each row of $n$ numbers forms an arithmetic sequence. We can directly use the sum formula for an arithmetic sequence to find the sum of each row of $n$ numbers, and then add the sums of the $n$ rows.
Solution: The sum of the numbers in the first row is $1+2+3+\cdots+n=\frac{n(n+1)}{2}$
The sum of the numbers in the second row is $2+3+4+\cdots+(n+1)=\frac{n(n+3)}{2}$
The sum of the numbers in the third row is $3+4+5+\cdots+(n+2)=\frac{n(n+5)}{2}$

The sum of the numbers in the $n$-th row is $n+(n+1)+(n+2)+\cdots+(2 n-1)=\frac{n(3 n-1)}{2}$
Adding up the above $n$ sums, we get the sum of the $n^{2}$ numbers in the given matrix:
$$
\begin{aligned}
S_{n} & =\frac{n(n+1)}{2}+\frac{n(n+3)}{2}+\frac{n(n+5)}{2}+\cdots+\frac{n(3 n-1)}{2} \\
& =\frac{n}{2}[(n+1)+(n+3)+(n+5)+\cdots+(3 n-1)] \\
& =\frac{n}{2} \cdot \frac{n[(n+1)+(3 n-1)]}{2}=n^{3}
\end{aligned}
$$