1. In a $4 \times 4$ grid, fill each cell with a different positive integer such that in each row and each column, exactly one cell contains a number that is the sum of the numbers in the other three cells. If the largest of these 16 positive integers is $M$, find the minimum value of $M$.
1. Let $a_{i}$ be the maximum number in the $i$-th column $(i=1,2,3,4)$, and let $b_{1}, b_{2}, \cdots, b_{12}$ be the remaining 12 numbers in the table.
Then, according to the problem,
$$
a_{1}+a_{2}+a_{3}+a_{4}=b_{1}+b_{2}+\cdots+b_{12} \text {, }
$$

where $b_{1}, b_{2}, \cdots, b_{12}$ are different positive integers.
$$
\begin{array}{l}
\text { Hence } a_{1}+a_{2}+a_{3}+a_{4}=b_{1}+b_{2}+\cdots+b_{12} \\
\geqslant 1+2+\cdots+12=78 .
\end{array}
$$

Since $a_{1}, a_{2}, a_{3}, a_{4}$ are four different positive integers, and $M$ is the largest among them, we have
$$
\begin{array}{l}
a_{1}+a_{2}+a_{3}+a_{4} \\
\leqslant M+(M-1)+(M-2)+(M-3) \\
=4 M-6 \\
\Rightarrow 4 M-6 \geqslant 78 \\
\Rightarrow M \geqslant 21 .
\end{array}
$$

This indicates that the minimum value of $M$ is 21, as shown in Figure 2.