For the set $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$ consisting of four different positive integers, define $S_{A}=a_{1}+a_{2}+a_{3}+a_{4}$. Let $n_{A}$ be the number of pairs $(i, j), 1 \leq i<j \leq 4$, such that $\left(a_{i}+a_{j}\right) \mid S_{A}$. Find all sets $A$ such that $n_{A}$ reaches its maximum value.
Third question,
Solution: Let's assume $a_{1}2\left(a_{2}+a_{4}\right)>S_{A}>a_{3}+a_{4}>a_{2}+a_{4}$, then $\frac{s_{A}}{a_{2}+a_{4}} \notin Z, \frac{s_{A}}{a_{3}+a_{4}} \notin Z$, hence $\mathrm{n}_{\mathrm{A}} \leq 4$.

If $n_{A}=4$, then $\frac{S_{A}}{a_{1}+a_{2}} 、 \frac{S_{A}}{a_{1}+a_{3}} 、 \frac{S_{A}}{a_{1}+a_{4}} 、 \frac{S_{A}}{a_{2}+a_{3}}$ are all positive integers. And $1a_{2}+ \left(a_{2}+a_{3}\right)=a_{2}+\left(a_{1}+a_{4}\right)>a_{2}+a_{4}>a_{1}+a_{3}$, so it must be that $a_{2}+a_{4}=2\left(a_{1}+a_{3}\right)$.

Combining $\left\{\begin{array}{l}a_{1}+a_{4}=a_{2}+a_{3} \\ a_{2}+a_{4}=2\left(a_{1}+a_{3}\right)\end{array}\right.$, we get $\left\{\begin{array}{l}a_{3}=2 a_{2}-3 a_{1} \\ a_{4}=3 a_{2}-4 a_{1}\end{array}\right.$. Also, note that $\frac{a_{3}+a_{4}}{a_{1}+a_{2}}>\frac{a_{2}+a_{4}}{a_{1}+a_{3}}=2$, and $\frac{a_{3}+a_{4}}{a_{1}+a_{2}}= \frac{\left(2 a_{2}-3 a_{1}\right)+\left(3 a_{2}-4 a_{1}\right)}{a_{1}+a_{2}}=\frac{5 a_{2}-7 a_{1}}{a_{1}+a_{2}}<5$, so it can only be that $\frac{5 a_{2}-7 a_{1}}{a_{1}+a_{2}}=3$ or 4.

If $\frac{5 a_{2}-7 a_{1}}{a_{1}+a_{2}}=3$, then $a_{2}=5 a_{1}$, and thus $\left\{\begin{array}{l}a_{3}=7 a_{1} \\ a_{4}=11 a_{1}\end{array}\right.$; if $\frac{5 a_{2}-7 a_{1}}{a_{1}+a_{2}}=4$, then $a_{2}=11 a_{1}$, and thus $\left\{\begin{array}{l}a_{3}=19 a_{1} \\ a_{4}=29 a_{1}\end{array}\right.$

In summary, the maximum value of $n_{A}$ is 4, and when $\mathrm{n}_{\mathrm{A}}=4$, $\mathrm{A}=\{\mathrm{c}, 5 c, 7 c, 11 c\}$ or $\mathrm{A}=\{c, 11 \mathrm{c}, 19 \mathrm{c}, 29 \mathrm{c}\}$, where $\mathrm{c}$ is any positive integer.