Example 5 In a $2 n \times 2 n$ grid, each cell is filled with the real number 1 or -1, such that exactly $2 n^{2}$ cells are filled with 1 and $2 n^{2}$ cells are filled with -1. $M$ is the minimum value of the absolute values of all row sums (the sum of all numbers in the same row) and column sums (the sum of all numbers in the same column). Determine the maximum possible value of $M$.
【Analysis】Since the absolute value of each row sum and column sum is not less than $M$, this problem needs to consider classification operations based on positive and negative values. By taking all the small squares at the intersection of positive (negative) rows and positive (negative) columns, the table is divided into four parts, with the sum of all numbers in each part being $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$, respectively.

This problem uses two different perspectives to estimate $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$: On one hand, all numbers in the table are 1 or -1, which can lead to an estimate of the operation of $S_{i}$. On the other hand, from the perspective of summing positive rows and positive columns, and summing negative rows and negative columns, another estimate of the operation of $S_{i}$ can be obtained. In summary, an inequality about $M$ can be derived, thereby obtaining an upper bound for $M$.

Solution: From the given information, the total sum of all row sums and column sums in the grid is 0. Suppose that among all row sums and column sums, there are $k$ rows and $l$ columns with non-negative values. By the pigeonhole principle, we can assume that $k+l \geqslant 2 n$.
Since row and column exchanges do not affect the structure of this problem, we can assume that in this grid, the first $k$ rows from top to bottom have non-negative row sums; the first $j$ columns from left to right have non-negative column sums.

On one hand, the first $k$ row sums and the first $l$ column sums are all non-negative, thus,
$$
\begin{array}{l}
\sum_{i=1}^{k} \sum_{j=1}^{2 n} a_{i j} \geqslant k M, \\
\sum_{i=1}^{2 n} \sum_{j=1}^{l} a_{i j} \geqslant l M, \\
\sum_{i=1}^{2 n} \sum_{j=1}^{2 n} a_{i j}=0 . \\
\text { Then } \sum_{i=1}^{k} \sum_{j=1}^{2 n} a_{i j}+\sum_{i=1}^{2 n} \sum_{j=1}^{l} a_{i j}-\sum_{i=1}^{2 n} \sum_{j=1}^{2 n} a_{i j} \\
\geqslant k M+l M, \\
\sum_{i=1}^{k} \sum_{j=1}^{l} a_{i j}-\sum_{i=k+1 j}^{2 n} \sum_{j=l+1}^{2 n} a_{i j} \geqslant l M+k M .
\end{array}
$$

That is, $\sum_{i=1}^{k} \sum_{j=1}^{l} a_{i j}-\sum_{i=k+1}^{2 n} \sum_{j=l+1}^{2 n} a_{i j} \geqslant l M+k M$.
On the other hand,
$$
\begin{array}{l}
\sum_{i=1}^{k} \sum_{j=1}^{l} a_{i j}-\sum_{i=k+1}^{2 n} \sum_{j=l+1}^{2 n} a_{i j} \\
\leqslant k l+(2 n-k)(2 n-l),
\end{array}
$$

Then $k l+(2 n-k)(2 n-l) \geqslant(k+l) M$, that is
$$
\begin{array}{l}
M \leqslant \frac{4 n^{2}-2(k+l) n+2 k l}{k+l} \\
\leqslant \frac{4 n^{2}+\frac{(k+l)^{2}}{2}}{k+l}-2 n .
\end{array}
$$

Therefore, $M \leqslant \frac{4 n^{2}}{k+l}+\frac{k+l}{2}-2 n, k+l \geqslant 2 n$.
Thus, $M \leqslant n$.
When $n$ is even, $M \leqslant n$;
When $n$ is odd, all row sums and column sums are even, at this time, $M \leqslant n-1$.

Specific construction: Divide the $2 n \times 2 n$ grid into four parts, the upper left $n \times n$ grid is filled with 1, the lower right $n \times n$ grid is filled with -1, the remaining two $n \times n$ grids are filled alternately with -1 and 1, ensuring that no two identical numbers are adjacent (two small squares with a common edge are considered adjacent).