For a given natural number $n \geqslant 2$, find the largest $N$ such that: no matter how the numbers $1,2, \cdots, n^{2}$ are filled into an $n \times n$ grid, there always exist two numbers in the same row or the same column whose difference is not less than $N$.

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Please note that the mathematical symbols and expressions are kept as they are, as they are universally understood and do not require translation.
The maximum value of $N$ is $\frac{(n-1)(n+2)}{2}$.
First, construct specific examples to prove:
$$
N \leqslant \frac{(n-1)(n+2)}{2} \text {. }
$$
(1) When $n$ is even, let $n=2 k$. As shown in Figure 2, divide the grid into four $k \times k$ regions $A, B, C, D$, and fill the numbers as follows.
The filling method for region $A$ is:
\begin{tabular}{cccc}
1 & 2 & $\cdots$ & $k$ \\
$k+1$ & $k+2$ & $\cdots$ & $2 k$ \\
$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
$k(k-1)+1$ & $k(k-1)+2$ & $\cdots$ & $k^{2}$
\end{tabular}

The filling method for region $B$ is to add $k^{2}$ to each number in region $A$, and similarly fill regions $C$ and $D$, adding $2 k^{2}$ and $3 k^{2}$ respectively.
(2) When $n$ is odd, let $n=2 k+1$. As shown in Figure 3, the middle row and middle column divide the rest of the grid into four $k \times k$ regions $A, B, C, D$, and fill the numbers as follows.
The filling method for region $A$ is the same as when $n$ is even, and the filling methods for regions $B, C, D$ are to add
$$
k^{2}+2 k, 2 k^{2}+2 k+1, 3 k^{2}+4 k+1 ;
$$

The numbers in the middle row from left to right are
$$
\begin{array}{r}
k^{2}+1, k^{2}+2, \cdots, k^{2}+k-1, k^{2}+k, 2 k^{2}+2 k+1, \\
3 k^{2}+2 k+2,3 k^{2}+2 k+3, \cdots, 3 k^{2}+3 k, 3 k^{2}+3 k+1 ;
\end{array}
$$

The numbers in the middle column from top to bottom are
$$
\begin{array}{l}
\quad k^{2}+k+1, k^{2}+k+2, \cdots, k^{2}+2 k-1, k^{2}+2 k, \\
2 k^{2}+2 k+1,3 k^{2}+3 k+2,3 k^{2}+3 k+3, \cdots, 3 k^{2}+4 k, \\
3 k^{2}+4 k+1 .
\end{array}
$$

It is easy to verify that for the above filling method, the difference between any two numbers in the same row or column of the grid does not exceed $\frac{(n-1)(n+2)}{2}$.

Secondly, for any filling method, let $1,2, \cdots,\left[\frac{n^{2}}{4}\right]+1$ occupy $x$ rows and $y$ columns, i.e., these $\left[\frac{n^{2}}{4}\right]+1$ numbers are all at the intersections of these $x$ rows and $y$ columns. Thus, $x y \geqslant\left[\frac{n^{2}}{4}\right]+1$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$).
$$
\begin{array}{l}
\text { Then } \frac{(x+y)^{2}}{4} \geqslant x y \geqslant\left[\frac{n^{2}}{4}\right]+1 \geqslant \frac{n^{2}+1}{4}>\frac{n^{2}}{4} \\
\Rightarrow x+y>n \Rightarrow x+y \geqslant n+1 .
\end{array}
$$

Let $n^{2}-\left[\frac{(n-1)^{2}}{4}\right], n^{2}-\left[\frac{(n-1)^{2}}{4}\right]+1, \cdots, n^{2}$
occupy $z$ rows and $w$ columns. Similarly,
$$
\begin{array}{l}
\frac{(z+w)^{2}}{4} \geqslant z w \geqslant\left[\frac{(n-1)^{2}}{4}\right]+1 \\
\geqslant \frac{n^{2}-2 n+2}{4}>\frac{(n-1)^{2}}{4} \\
\Rightarrow z+w>n-1 \Rightarrow z+w \geqslant n .
\end{array}
$$

From equations (1) and (2),
$$
x+y+z+w \geqslant 2 n+1>2 n \text {. }
$$

Therefore, there exists a number $u$ in $1,2, \cdots,\left[\frac{n^{2}}{4}\right]+1$ and a number $v$ in
$$
n^{2}-\left[\frac{(n-1)^{2}}{4}\right], n^{2}-\left[\frac{(n-1)^{2}}{4}\right]+1, \cdots, n^{2}
$$

in the same row or column, and
$$
\begin{array}{l}
v-u \geqslant\left(n^{2}-\left[\frac{(n-1)^{2}}{4}\right]\right)-\left(\left[\frac{n^{2}}{4}\right]+1\right) \\
=n^{2}-\left[\frac{(n-1)^{2}}{4}+\frac{n^{2}}{4}\right]-1 \\
=n^{2}-\frac{2 n^{2}-2 n}{4}-1=\frac{(n-1)(n+2)}{2}
\end{array}
$$
(Note that, $\frac{(n-1)^{2}}{4}$ and $\frac{n^{2}}{4}$ must have one as an integer).
Thus, $N$ satisfies the requirement.
In summary, the maximum value of $N$ is $\frac{(n-1)(n+2)}{2}$.