10.8 Try to find the largest positive integer $N$, such that no matter how the positive integers from 1 to 400 are filled into the cells of a $20 \times 20$ grid, one can always find two numbers in the same row or the same column whose difference is not less than $N$.
10.8209.

First, we illustrate that $N \leqslant 209$. Divide the grid into two $20 \times 10$ grids using a vertical line through the center. Fill the numbers from 1 to 200 into the left grid in increasing order row by row, and fill the numbers from 201 to 400 into the right grid in the same manner. This way, the maximum difference between the numbers in each row does not exceed $210-1=209$; the maximum difference between the numbers in each column does not exceed $191-1=190$. Therefore, $N \leqslant 209$.
Next, we prove that $N$ cannot be less than 209. Consider the subsets
$$
M_{1}=\{1,2, \cdots, 91\} \text { and } M_{2}=\{300,301, \cdots, 400\} \text {. }
$$

Color the rows and columns that contain numbers from $M_{1}$ in red; color the rows and columns that contain numbers from $M_{2}$ in blue. We need to show that the number of red rows and columns is at least 20, and the number of blue rows and columns is at least 21. If this is true, then there must be a row or column that is both red and blue, meaning there are two numbers in it with a difference of at least $300-91=209$.

Suppose there are $i$ rows and $j$ columns colored red. Then, all elements of $M_{1}$ are located at the intersections of these rows and columns, so $i j \geqslant 91$.
Thus, $i+j \geqslant 2 \sqrt{\ddot{j}} \geqslant 2 \sqrt{91} \geqslant 19$.
Similarly, the sum of the number of blue rows and columns
$$
i^{\prime}+j^{\prime} \geqslant 2 \sqrt{i^{\prime} j^{\prime}} \geqslant 2 \sqrt{101}>20 .
$$