Given a height of 624, find the largest positive integer $n$ such that when the positive integers from 1 to $m^{2}\left(m \geqslant 2, m \in \mathbf{Z}_{+}\right)$ are arbitrarily filled into the $m \times m$ grid of $m^{2}$ cells, there will always be a row or a column where the difference between two numbers is not less than $n$.
(1) When $m$ is even, i.e., $m=2 k\left(k \in \mathbf{Z}_{+}\right)$, let
$$
\begin{array}{l}
M=\left\{1,2, \cdots, k^{2}+1\right\}, \\
N=\left\{3 k^{2}+k, 3 k^{2}+k+1, \cdots, 4 k^{2}\right\} .
\end{array}
$$
$$
\text { Then }|M|=k^{2}+1,|N|=k^{2}-k+1 \text {. }
$$

Assume the table is filled arbitrarily, and there are $i_{1}$ rows and $j_{1}$ columns containing numbers from $M$, and $i_{2}$ rows and $j_{2}$ columns containing numbers from $N$. Then
$$
\begin{array}{l}
i_{1}+j_{1} \geqslant 2 \sqrt{i_{1} j_{1}} \geqslant 2 \sqrt{|M|}=2 \sqrt{k^{2}+1}>2 k \\
\Rightarrow i_{1}+j_{1} \geqslant 2 k+1 .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
i_{2}+j_{2} \geqslant 2 \sqrt{i_{2} j_{2}} \geqslant 2 \sqrt{|N|}=2 \sqrt{k^{2}-k+1}>2 k-1 \\
\Rightarrow i_{2}+j_{2} \geqslant 2 k . \\
\text { Hence } i_{1}+j_{1}+i_{2}+j_{2} \geqslant 4 k+1>4 k .
\end{array}
$$

Using the pigeonhole principle, we know there must be a row or a column that contains one number from each of the two sets, and the difference between these two numbers is at least $2 k^{2}+k-1$.
Thus, $n_{\max } \geqslant 2 k^{2}+k-1$.
Furthermore, when the numbers are filled as shown in Figure 3, the difference between any two numbers in the same row is no more than $2 k^{2}+k-1$, and the difference between any two numbers in the same column is no more than $2 k^{2}-k$, so $n \leqslant 2 k^{2}+k-1$.
$$
\begin{array}{cccccccc}
1 & 2 & \cdots & k & 2 k^{2}+1 & 2 k^{2}+2 & \cdots & 2 k^{2}+k \\
k+1 & k+2 & \cdots & 2 k & 2 k^{2}+k+1 & 2 k^{2}+k+2 & \cdots & 2 k^{2}+2 k \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
2 k^{2}-k+1 & 2 k^{2}-k+2 & \cdots & 2 k^{2} & 4 k^{2}-k+1 & 4 k^{2}-k+2 & \cdots & 4 k^{2}
\end{array}
$$

Figure 3
$$
\text { In summary, } n_{\max }=2 k^{2}+k-1=\frac{m^{2}+m-2}{2} \text {. }
$$
(2) When $m$ is odd, i.e., $m=2 k+1\left(k \in \mathbf{Z}_{+}\right)$, let
$$
\begin{aligned}
M= & \left\{1,2, \cdots, k^{2}+1\right\}, \\
N= & \left\{3 k^{2}+3 k+1,3 k^{2}+3 k+2, \cdots,\right. \\
& \left.(2 k+1)^{2}\right\} .
\end{aligned}
$$

Then $|M|=k^{2}+1,|N|=k^{2}+k+1$.
Assume the table is filled arbitrarily, and there are $i_{1}$ rows and $j_{1}$ columns containing numbers from $M$, and $i_{2}$ rows and $j_{2}$ columns containing numbers from $N$. Similarly to (1),
$$
\begin{array}{l}
i_{1}+j_{1} \geqslant 2 \sqrt{i_{1} j_{1}} \geqslant 2 \sqrt{|M|} \\
=2 \sqrt{k^{2}+1}>2 k \\
\Rightarrow i_{1}+j_{1} \geqslant 2 k+1, \\
i_{2}+j_{2} \geqslant 2 \sqrt{i_{2} j_{2}} \geqslant 2 \sqrt{|N|} \\
=2 \sqrt{k^{2}+k+1}>2 k+1 \\
\Rightarrow i_{2}+j_{2} \geqslant 2 k+2 . \\
\text { Hence } i_{1}+j_{1}+i_{2}+j_{2} \geqslant 4 k+3>4 k+2 .
\end{array}
$$

Using the pigeonhole principle, we know there must be a row or a column that contains one number from each of the two sets, and the difference between these two numbers is at least $2 k^{2}+3 k$.
Thus, $n_{\max } \geqslant 2 k^{2}+3 k$.
Let Figure 4 be a “$k$-order standard grid table”.
\begin{tabular}{cccc}
1 & 2 & $\cdots$ & $k$ \\
$k+1$ & $k+2$ & $\cdots$ & $2 k$ \\
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\
$k^{2}-k+1$ & $k^{2}-k+2$ & $\cdots$ & $k^{2}$
\end{tabular}

Figure 4
In Figure 5, fill the four regions I, II, III, and IV with 4 $k$-order standard grid tables, then add $2 k^{2}+2 k+1$ to the numbers in region II, add $k^{2}+2 k$ to the numbers in region III, and add $3 k^{2}+4 k+1$ to the numbers in region IV.

Filling the numbers as shown in Figure 5, we get that the difference between any two numbers in the same row or column is no more than $2 k^{2}+3 k$, so
$$
n \leqslant 2 k^{2}+3 k \text {. }
$$
$\square$
$\square$
$$
k^{2}+2 k
$$
$\square$ $3 k^{2}+3 k+2$ $3 k^{2}+4 k+1$
IV
Figure 5

In summary, $n_{\text {max }}=2 k^{2}+3 k=\frac{m^{2}+m-2}{2}$.
Combining (1) and (2), we have $n_{\max }=\frac{m^{2}+m-2}{2}$.