6. Determine the maximum value of the integer $N$ that satisfies the following conditions: In a $5 \times 5$ grid, fill in the numbers from $1 \sim 25$ such that each cell contains a different number, and the sum of the numbers in some $2 \times 2$ subgrid is at least $N$.
6. The maximum value sought is $N=45$, an example is shown in Figure 2.

If we let $n=5, m=\left\lceil\frac{n}{2}\right\rceil=3$, we will next prove: if $n$ is odd, in an $n \times n$ grid, fill in $1 \sim n^{2}$, then the sum of the numbers in some $2 \times 2$ subgrid is at least $8 m^{2}-11 m+6$.

Let the number filled in the cell at the $i$-th row and $j$-th column be $a_{i j}$, and the sum of the numbers in the $2n-1$ cells of the $i$-th row and $j$-th column be $s_{i j}$. Then,
$$
s_{i j}=\sum_{k=1}^{n}\left(a_{i k}+a_{k j}\right)-a_{i j} \text {. }
$$

Consider the sum $S$
$$
\begin{array}{l}
=\sum_{\substack{i=1 \\
j=1}}^{m} s_{2 i-1,2 j-1}=\sum_{i=j=1(\bmod 2)} s_{i j} \\
=(2 m-1) \sum_{i=j=1(\bmod 2)} a_{i j}+m \sum_{i+j=1(\bmod 2)} a_{i j},
\end{array}
$$

where there are $m^{2}$ ordered pairs $(i, j)$ satisfying
$$
i \equiv j \equiv 1(\bmod 2),
$$
and $2 m(n-m)$ ordered pairs $(i, j)$ satisfying
$$
i+j \equiv 1(\bmod 2) \text {. }
$$

Clearly, the maximum value that $S$ can achieve is
$$
\begin{array}{l} 
S_{0}=(2 m-1) \sum_{k=0}^{m^{2}-1}\left(n^{2}-k\right)+ \\
m \sum_{k=0}^{2 m(n-m)-1}\left(n^{2}-m^{2}-k\right) .
\end{array}
$$

Thus, $s_{k l} \leqslant \frac{S_{0}}{m^{2}}$ holds for some odd indices $k, l$.
In this problem,
$$
\begin{array}{l} 
S_{0}= 5 \times(25+24+\cdots+17)+ \\
3 \times(16+15+\cdots+5) \\
=1323 .
\end{array}
$$
$$
\text { Hence } s_{k l} \leqslant \frac{1323}{9}=147 \text {. }
$$
In the $n \times n$ grid, the sum of the numbers in the remaining cells after removing the $2n-1$ cells in the $k$-th row and $l$-th column is at least
$$
\left(1+2+\cdots+n^{2}\right)-\frac{S_{0}}{m^{2}}=\frac{n^{2}\left(n^{2}+1\right)}{2}-\frac{S_{0}}{m^{2}},
$$

In this problem, it is at least
$$
(1+2+\cdots+25)-147=178 \text {. }
$$

Finally, since $n$ is odd, $n=2 m-1$, so the remaining cells can be divided into $(m-1)^{2}$ $2 \times 2$ subgrids, indicating that the sum of the numbers in at least one $2 \times 2$ subgrid is at least
$$
\begin{array}{l}
\left\lceil\frac{n^{2}\left(n^{2}+1\right)}{2(m-1)^{2}}-\frac{S_{0}}{m^{2}(m-1)^{2}}\right\rceil \\
=\left\lceil 8 m^{2}-11 m+5+\frac{1}{2}\right\rceil \\
=8 m^{2}-11 m+6 .
\end{array}
$$

Therefore, in this problem, the result is
$$
8 \times 3^{2}-11 \times 3+6=45 \text {. }
$$