I saw this problem on a maths challenge book. Given any set $A=\{a_1 ,a_2, a_3, a_4\}$ of four distinct positive integers, we denote the sum $a_1+a_2+a_3+a_4$ by $S\{A\}$. Let $n_A$ denote the number of pairs $(i,j)$ with $1\le i\le j\le4$ for which $a_i +a_j$ divides $S\{A\}$. Find all sets of four distinct positive integers which achieve the largest value of $n_A$.
I was thinking that ${4\choose 2}$ is $6$. So there should be six possible ways to pair all the elements of set $A$ that will divide $S\{A\}$ provided that $a_i + a_i$ is not allowed. $n_A$ will have the value of $6$ if what I said is possible. Well this is only my first thought, other ideas will be appreciated .
WLOG assume that $a_1<a_2<a_3<a_4$. Then $$a_3+a_4>a_1+a_2 \implies a_3+a_4>\frac{1}{2}S \implies (a_3+a_4)\nmid S$$ and $$a_2+a_4>a_1+a_3 \implies a_2+a_4>\frac{1}{2}S \implies (a_2+a_4)\nmid S$$ So two of the six possible combinations definitely don't divide the sum. So $n_A$ is at most four.
We have maximal $n_A=4$, because there is a solution for $n_A=4$, namely
$$(a_1,a_2,a_3,a_4)=k(1,5,7,11),\quad k\in\mathbb{Z}^+$$
This was constructed by trial and error after noting that if $n_A=4$ then:
So first try with multiples of $2$, $3$ and $4$, resulting in the set of simultaneous equations
$$\begin{array}{rcccc} S=&2a_1&&&+2a_4 \\ S=&&2a_2&+2a_3 \\ S=&3a_1&&+3a_3 \\ S=&4a_1&+4a_2 \\ \end{array}$$
This solves to $a_1=\frac{S}{24},a_2=\frac{5S}{24},a_3=\frac{7S}{24},a_4=\frac{11S}{24}$.
Since the $a_i$ must be integral, take $S=24k,a_1=k,a_2=5k,a_3=7k,a_4=11k$.
I will leave it up to the reader to see whether multiples other than $3$ and $4$ can also produce viable solutions, noting that solutions are not viable if one of the numbers (especially $a_1$) is a negative multiple of $S$.