14. Find the largest positive integer $n$, such that when the positive integers from 1 to 400 are arbitrarily filled into the 400 cells of a $20 \times 20$ grid, there will always be a row or a column where the difference between two numbers is not less than $n$.
Divide the square table into two $20 \times 10$ rectangular tables, then fill the numbers from 1 to 200 into the left table in ascending order row by row, and fill the numbers from 201 to 400 into the right table in the same manner. In this table, the difference between any two numbers in the same row does not exceed 209, and the difference between any two numbers in the same column does not exceed 190, so $n \leqslant 209$.

Let $M=\{1,2, \cdots, 91\}, N=\{300,301, \cdots, 400\}$. We will prove that there must be a row or a column that contains a number from both sets $M$ and $N$.

Assume the numbers in set $M$ are distributed in $i$ rows and $j$ columns, and the numbers in set $N$ are distributed in $k$ rows and $l$ columns. Thus, $i+j \geqslant 2 \sqrt{i j} \geqslant 2 \sqrt{91}>19, k+l \geqslant 2 \sqrt{k l} \geqslant 2 \sqrt{101}>20$, which implies $i+j \geqslant 20, k+l \geqslant 21 \Rightarrow i+j+k+l \geqslant 41$. Therefore, there must be a row or a column that contains a number from both sets $M$ and $N$, and the difference between these two numbers is at least 209. Hence, the conclusion holds.

In summary, the maximum value of $n$ is 209.