Firstly, we will prove that the maximum value of $p_{A}$ is at most 4. Without loss of generality, we may assume that $a_{1}a_{1}+a_{3}$ and $a_{3}+a_{4}>a_{1}+a_{2}$. Hence $a_{2}+a_{4}$ and $a_{3}+a_{4}$ do not divide $s_{A}$. This proves $p_{A} \leq 4$. Now suppose $p_{A}=4$. By the previous argument we have
$$\begin{array}{lll} a_{1}+a_{4} \mid a_{2}+a_{3} & \text { and } & a_{2}+a_{3} \mid a_{1}+a_{4}, \\ a_{1}+a_{2} \mid a_{3}+a_{4} & \text { and } & a_{3}+a_{4} \not a_{1}+a_{2}, \\ a_{1}+a_{3} \mid a_{2}+a_{4} & \text { and } & a_{2}+a_{4} \not a_{1}+a_{3}. \end{array}$$
Hence, there exist positive integers $m$ and $n$ with $m>n \geq 2$ such that
$$\left\{\begin{array}{l} a_{1}+a_{4}=a_{2}+a_{3} \\ m\left(a_{1}+a_{2}\right)=a_{3}+a_{4} \\ n\left(a_{1}+a_{3}\right)=a_{2}+a_{4} \end{array}\right.$$
Adding up the first equation and the third one, we get $n\left(a_{1}+a_{3}\right)=2 a_{2}+a_{3}-a_{1}$. If $n \geq 3$, then $n\left(a_{1}+a_{3}\right)>3 a_{3}>2 a_{2}+a_{3}>2 a_{2}+a_{3}-a_{1}$. This is a contradiction. Therefore $n=2$. If we multiply by 2 the sum of the first equation and the third one, we obtain
$$6 a_{1}+2 a_{3}=4 a_{2}$$
while the sum of the first one and the second one is
$$(m+1) a_{1}+(m-1) a_{2}=2 a_{3}.$$
Adding up the last two equations we get
$$(m+7) a_{1}=(5-m) a_{2}.$$
It follows that $5-m \geq 1$, because the left-hand side of the last equation and $a_{2}$ are positive. Since we have $m>n=2$, the integer $m$ can be equal only to either 3 or 4. Substituting $(3,2)$ and $(4,2)$ for $(m, n)$ and solving the previous system of equations, we find the families of solutions $\{d, 5 d, 7 d, 11 d\}$ and $\{d, 11 d, 19 d, 29 d\}$, where $d$ is any positive integer.