1. For any set $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$ consisting of four different positive integers, let $S_{A}=a_{1}+a_{2}+a_{3}+a_{4}$. Let $n_{A}$ be the number of pairs $(i, j)$ such that $a_{i}+a_{j}(1 \leqslant i<j \leqslant 4)$ divides $S_{1}$. Find all sets $A$ consisting of four different positive integers such that $n_{A}$ reaches its maximum value.
For a set of four positive integers
$$
A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\} \text {, }
$$

without loss of generality, assume $a_{1} < a_{2} < a_{3} < a_{4}$.
Let $S_{1}=a_{1}+a_{2}+a_{3}+a_{4}$, and $u=\operatorname{gcd}\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$.
Then, $u \mid S_{1}$.
Since $a_{1}+a_{2} \mid S_{1}$, $a_{1}+a_{2} \mid a_{3}+a_{4}$.
Similarly, $a_{1}+a_{3} \mid a_{2}+a_{4}$.
Thus, $a_{1}+a_{2} \mid a_{4}-a_{2}$ and $a_{1}+a_{3} \mid a_{4}-a_{2}$.
Since $a_{1}+a_{2} < a_{1}+a_{3} < a_{4}-a_{2}$, we have $a_{4}-a_{2} \geq 2\left(a_{1}+a_{2}\right)$.
Therefore, $a_{4} \geq 3 a_{2} > 3 a_{1}+4 a_{1}$.
Thus, $u=6 a_{1}$ or $12 a_{1}$.
Hence, $A=\left\{a_{1}, 5 a_{1}, 7 a_{1}, 11 a_{1}\right\}$
or $A=\left\{a_{1}, 11 a_{1}, 19 a_{1}, 29 a_{1}\right\}$.
It is easy to verify that the above sets all have
$$
a_{1}+a_{2} 、 a_{1}+a_{3} 、 a_{1}+a_{4} 、 a_{2}+a_{3}
$$

dividing $S_{1}$.
In summary, when $n_{A}$ reaches its maximum value of 4,
$$
A=\{a, 5 a, 7 a, 11 a\}
$$

or $A=\{a, 11 a, 19 a, 29 a\}$,
where $a$ is a positive integer.