Example 4 Figure 1 is a chessboard that can extend infinitely to the right and downward, with horizontal rows and vertical columns. Fill in the squares with positive integers according to the pattern shown in the $4 \times 4$ grid already filled.
(1) Find the number in the square located at the 3rd row and 8th column;
(2) In which square is the number 321?
(3) Write the general term formula for the sequence of numbers in the squares along the diagonal from the top-left corner to the bottom-right corner.
(4) Find the sum of the first $n$ terms, $S_{n}$, of the sequence in (3).
Explanation: Grouping diagonally from top-right to bottom-left:
(1), (2,3), (4,5,6), ,

The $k$-th group has $k$ numbers, and the column of the first number in this group is the $k$-th column. The first $k-1$ groups have
$$
1+2+3+\cdots+(k-1)=\frac{k(k-1)}{2}
$$

numbers.
Let the number in the $i$-th row and $j$-th column be $a_{ij}$, which is in the $k$-th group. Starting from $a_{ij}$, moving up one number in the direction of the arrow increases the column $j$ by 1. Moving up $i-1$ numbers gives the first number of the group, so $k=j+i-1$. $a_{ij}$ is the $i$-th number in the $j+i-1$ group, which is equal to the last number of the $j+i-2$ group plus $i$, i.e.,
$$
a_{ij}=\frac{(i+j-2)(i+j-1)}{2}+i.
$$
(1) Substituting $i=3, j=8$ into (1) gives the number in the 3rd row and 8th column as 48.
(2) Suppose 321 is in the $k$-th group, then
$$
\frac{k(k-1)}{2}+1 \leqslant 321<\frac{k(k+1)}{2}+1.
$$

Solving this gives $k=25$.
$$
\text{Hence }\left\{\begin{array}{l}
j+i-1=25, \\
a_{ij}=\frac{(i+j-2)(i+j-1)}{2}+i=321.
\end{array}\right.
$$

Solving this gives $i=21, j=5$.
Therefore, 321 is located in the 21st row and 5th column.
(3) Let the sequence satisfying the condition be $\left\{b_{n}\right\}$. $b_{n}$ is the middle number of the $2n-1$ group, and the first and last numbers of this group are
$$
\begin{array}{l}
\frac{(2n-1)(2n-2)}{2}+1, \\
\frac{(2n-1)(2n-2)}{2}+2n-1.
\end{array}
$$

Since $b_{n}$ is the arithmetic mean of the two, we have
$$
b_{n}=2n^2-2n+1.
$$
(4) $S_{n}=2\left(1^2+2^2+\cdots+n^2\right)-2(1+2+\cdots+n)+n$
$$
=\frac{n\left(2n^2+1\right)}{3}.
$$