Let $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$ be a set of four distinct positive integers. We denote the sum $a_{1}+a_{2}+a_{3}+a_{4}$ by $s_{A}$, and let $n_{A}$ be the number of pairs $(i, j)$ $(1 \leq i<j \leq 4)$ for which $a_{i}+a_{j}$ divides $s_{A}$.

Determine all sets $A$ for which $n_{A}$ takes its maximum possible value.
Janzer Olivér's solution. By symmetry, we can assume that $a_{1}a_{1}+a_{3}$, so $a_{2}+a_{4}>\frac{1}{2} s_{A}$, thus it cannot be a divisor of $s_{A}$ (since $a_{2}+a_{4}a_{1}+a_{2}$, so $a_{3}+a_{4}>\frac{1}{2} s_{A}$, thus this cannot be a divisor either.

Therefore, only the pairs $\left(a_{1}, a_{2}\right),\left(a_{1}, a_{3}\right),\left(a_{1}, a_{4}\right),\left(a_{2}, a_{3}\right)$ can be considered, meaning at most 4 pairs. Let's see when this number is achieved. Since $\left(a_{1}+a_{4}\right)$ and $\left(a_{2}+a_{3}\right)$ are both divisors, $\left(a_{1}+a_{4}\right) \leq \frac{1}{2} s_{A}$ and $\left(a_{2}+a_{3}\right) \leq \frac{1}{2} s_{A}$, so $\left(a_{1}+a_{2}+a_{3}+a_{4}\right) \leq s_{A}$. Equality must hold: $\left(a_{1}+a_{4}\right)=\left(a_{2}+a_{3}\right)=\frac{1}{2} s_{A}$.

$\left(a_{1}+a_{3}\right)$ is also a divisor of $s_{A}$, so $\left(a_{1}+a_{3}\right) \leq \frac{1}{3} s_{A}$, since $\left(a_{1}+a_{3}\right)a_{2}$ leads to a contradiction. Thus $\frac{1}{4} s_{A}<\left(a_{1}+a_{3}\right) \leq \frac{1}{3} s_{A}$, so $\left(a_{1}+a_{3}\right)=\frac{1}{3} s_{A}$.

Since $a_{1}+a_{4}=a_{2}+a_{3}, a_{4}=a_{2}+a_{3}-a_{1}$

$$
\begin{aligned}
a_{1}+a_{3} & =\frac{1}{3} s_{A}=\frac{1}{3}\left(a_{1}+a_{2}+a_{3}+a_{4}\right)=\frac{1}{3}\left(a_{1}+a_{2}+a_{3}+\left(a_{2}+a_{3}-a_{1}\right)\right)= \\
& =\frac{1}{3}\left(2 a_{2}+2 a_{3}\right)
\end{aligned}
$$

Thus $3 a_{1}+3 a_{3}=2 a_{2}+2 a_{3}$, so $a_{3}=2 a_{2}-3 a_{1}$,

$$
a_{4}=a_{2}+a_{3}-a_{1}=a_{2}+\left(2 a_{2}-3 a_{1}\right)-a_{1}=3 a_{2}-4 a_{1}
$$

$\left(a_{1}+a_{2}\right)$ is also a divisor of $s_{A}$, but $\left(a_{1}+a_{2}\right)<\left(a_{1}+a_{3}\right)=\frac{1}{3} s_{A}$, so $\left(a_{1}+a_{2}\right) \leq \frac{1}{4} s_{A}$. Suppose $\left(a_{1}+a_{2}\right) \leq \frac{1}{6} s_{A}$. Then

$$
\left(a_{1}+a_{2}\right) \leq \frac{1}{6}\left(a_{1}+a_{2}+\left(2 a_{2}-3 a_{1}\right)+\left(3 a_{2}-4 a_{1}\right)\right)=\frac{1}{6}\left(6 a_{2}-6 a_{1}\right)=a_{2}-a_{1}
$$

which implies $2 a_{1} \leq 0$, a contradiction due to the positivity of $a_{1}$.

Thus $\frac{1}{6} s_{A}<\left(a_{1}+a_{2}\right) \leq \frac{1}{4} s_{A}$, leaving two cases:
1)

$$
a_{1}+a_{2}=\frac{1}{4} s_{A}=\frac{1}{4}\left(a_{1}+a_{2}+\left(2 a_{2}-3 a_{1}\right)+\left(3 a_{2}-4 a_{1}\right)\right)=\frac{1}{4}\left(6 a_{2}-6 a_{1}\right) .
$$

Thus $4 a_{1}+4 a_{2}=6 a_{2}-6 a_{1}$, so $10 a_{1}=2 a_{2}$, or $a_{2}=5 a_{1}$. Then $a_{3}=2 a_{2}-3 a_{1}=7 a_{1}, a_{4}=3 a_{2}-4 a_{1}=11 a_{1}$. These are valid if $a_{1}$ is a positive integer, since $s_{A}=24 a_{1}$, and $a_{1}+a_{2}=6 a_{1}, a_{1}+a_{3}=8 a_{1}, a_{1}+a_{4}=12 a_{1}, a_{2}+a_{3}=12 a_{1}$, which are indeed divisors.
2)

$$
a_{1}+a_{2}=\frac{1}{5} s_{A}=\frac{1}{5}\left(6 a_{2}-6 a_{1}\right), \quad 5 a_{1}+5 a_{2}=6 a_{2}-6 a_{1}
$$

$a_{2}=11 a_{1}, a_{3}=19 a_{1}, a_{4}=29 a_{1}$. Thus $s_{A}=60 a_{1}$ and $a_{1}+a_{2}=12 a_{1}, a_{1}+a_{3}=20 a_{1}, a_{1}+a_{4}=30 a_{1}, a_{2}+a_{3}=30 a_{1}$, which are indeed divisors.

Therefore, the sets $A$ that satisfy the conditions are those where $A=\{a, 5 a, 7 a, 11 a\}$ or $A=\{a, 11 a, 19 a, 29 a\}$ and $a$ is a positive integer.