14. (20 points) Find the largest positive integer $n$, such that when the positive integers from 1 to 400 are arbitrarily filled into a $20 \times 20$ grid, there will always be a row or a column where the difference between two numbers is not less than $n$.
14. First, divide the original grid into two rectangular grids of $20 \times 10$, then fill the numbers from 1 to 200 into the left grid in ascending order row by row. Similarly, fill the numbers from 201 to 400 into the right grid. In this grid, the difference between any two numbers in the same row is no more than 209, and the difference between any two numbers in the same column is no more than 190, hence $n \leqslant 209$.
$$
\begin{array}{l}
\text { Let } M=\{1,2, \cdots, 91\}, \\
N=\{300,301, \cdots, 400\} .
\end{array}
$$

It is easy to see that if there is a number from each of the sets $M$ and $N$ in the same row or column of the grid, then the difference between these two numbers is at least 209. Therefore, we only need to prove that in every row and every column of the grid, there must be a number from each of the sets $M$ and $N$.

Assume the grid is already filled arbitrarily, and there are $i$ rows and $j$ columns containing numbers from set $M$, and $k$ rows and $l$ columns containing numbers from set $N$. Then
$$
\begin{array}{l}
i+j \geqslant 2 \sqrt{i j} \geqslant 2 \sqrt{91}>19 \\
\Rightarrow i+j \geqslant 20 .
\end{array}
$$

Similarly, $k+l \geqslant 2 \sqrt{k l} \geqslant 2 \sqrt{101}>20$.
Thus, $i+j+k+l \geqslant 41$.
By the pigeonhole principle, there must be a row or column that contains a number from each of the two sets. The proposition is proved.
In conclusion, $n=209$.