I think it is easy to prove that \[ n_A \]<5. Hence: \[ n_A = 4 \]

Let $a_1<a_2<a_3<a_4$. We have $a_2+a_4>a_1+a_3$ and $a_3+a_4>a_1+a_2$. Therefore, $a_2+a_4$ and $a_3+a_4$ are greater then $\frac{s_A}{2}$ and do not divide $s_A$. So the number $n_A$ cannot be greater than 4. To achieve the maximum value, consider $a_1+a_4$ and $a_2+a_3$. If they are not equal, one of them has to be greater than $\frac{s_A}{2}$ and does not divide $s_A$. Therefore, $a_1+a_4 = a_2+a_3$. Let $k=a_2-a_1$, $m=a_1$, $n=a_4$. The four numbers are in the forms $m,m+k,n-k,n$. We have $2m+k | 2(m+n)$ and $m+n-k | 2(m+n)$. Since $a_2<a_3$, we get $m+k<n-k$. So, $2k<n-m<m+n$. Therefore, $4(m+n-k)>2(m+n)>2(m+n-k)$. But since $m+n-k | 2(m+n)$, the only case is $2(m+n)=3(m+n-k)$ or $k=\frac{m+n}{3}$. Now the four numbers are in the forms $m, \frac{4m+n}{3},\frac{2n-m}{3},n$. Since $a_1+a_2=\frac{7m+n}{3}$ divides $2(m+n)$, we get $7m+n | 6m+6n$ or $7m+n|36n$.

The answers are $(1,5,7,11),(1,11,19,29),(2,10,14,22)$ and their multiples. In fact, by applying several inequalities, we know that the answer has the form $(a,b,d-b,d-a)$ where $\frac{d}{2}>b>a,(a,b)=1$. To achieve the maximum value 4, we further need $d-(b-a)|2d$ and $a+b|2d$, but since $d-(b-a)>\frac{d}{2}$, we must have $3(a-b)=d$. All we need to do is to solve $a+b|3(b-a)$, which is pretty easy.

Yes, $\{(a,11a,19a,29a)|a\in\mathbb N\}$ or $\{(a,5a,7a,11a)|a\in\mathbb N\}$.

WLOG $a_1<a_2<a_3<a_4$. Let $a_1=a$, $a_2=a+x$, $a_3=a+x+y$, $a_4=a+x+y+z$ ($x,y,z>0$). We see that $a_4+a_3>a_1+a_2$ and $a_4+a_2>a_1+a_3$ so $s_A$ is not divisible by $a_4+a_3$, and $a_4+a_2$. So because we can choose $i,j$ on $6$ ways, and $2$ ways are impossible, $n_A\leq 4$. It is easy to see that $n=4$ because numbers $1,7,5,11$ satisfies condition. Now we have: $2a+x|2a+2x+2y+z\Rightarrow 2a+x|2(x+y)$ $2a+x+y|2a+3x+y+z\Rightarrow 2a+x+y|x+z$ $2a+x+y+z|2a+2x+y$....$[1]$ $2a+2x+y|2a+x+y+z$....$[2]$ $[1],[2]\Rightarrow 2a+x+y+z=2a+2x+y\Rightarrow x=z$ (because a|b and b|a) Now we just need to find numbers $a,x,y$ such that: $2a+x|2(x+y)$ and $2a+x+y|2x\Rightarrow k(2a+x+y)=2x \Rightarrow k=1$ because for $k>1$ $k(2a+x+y)>2x$, so we have $2a+y=x$. Now $2a+x=4a+y|2(x+y)=4a+y+3y$ so $4a+y|3y\Rightarrow m(4a+y)=3y$ so $m=1$ or $m=2$. If $m=1$ then solutions are $(a_1,a_2,a_3,a_4)=(a,7a,5a,11a)$ and if $m=2$ solutions are $(a_1,a_2,a_3,a_4)=(a,11a,19a,29a)$ and all permutations. note: I use that if $a_i+a_j|s_A\Rightarrow a_i+a_j|s_A-a_i-a_j$, and if $a|b$ then $a\leq b$.

it is very nice problem))) i solved it in about hour))) but the second and third are awful))

Suppose WLOG $a_1 < a_2 < a_3 < a_4$. It is not possible that $a_3 + a_4 \mid s_A$, because it is equivalent to $a_3 + a_4 \mid a_1 + a_2$, a contradiction since $a_1,a_2,a_3,a_4$ are positive and $a_3+a_4 > a_1+a_2$. In a similar fashion we can rule out the case $a_2+a_4 \mid s_A$. Therefore we get $n_A \leq 4$. If $n_A = 4$, the following conditions must be fulfilled: $a_1+a_2 \mid a_3+a_4$; $a_1+a_3 \mid a_2+a_4$; $a_1+a_4 \mid a_2+a_3$; $a_2+a_3 \mid a_1+a_4$. The last two conditions give $a_1+a_4 = a_2+a_3$, since the $a_i$'s are positive. Hence $s_A = 2(a_2+a_3)$, and now we just have to find all positive integers $a_1 < a_2 < a_3$ such that $a_1+a_2 \mid 2(a_2+a_3)$ and $a_1+a_3 \mid 2(a_2+a_3)$. Let $2(a_2+a_3) = k(a_1+a_2)$, hence $a_3 = \frac{ka_1+(k-2)a_2}{2}$. Notice $k \geq 3$ since $2(a_1+a_2) < 2(a_2+a_3)$. The second condition is equivalent to $a_1+a_3 \mid 2(a_2-a_1)$, that is, $\frac{(k+2)a_1+(k-2)a_2}{2} \mid 2(a_2-a_1) \iff (k+2)a_1+(k-2)a_2 \mid 4a_2 - 4a_1 \Rightarrow k \leq 5$, otherwise LHS would be greater than RHS. So now we have three cases: If $k=5$, then $7a_1+3a_2 \mid 4a_2 - 4a_1$, but we must have equality since $2(7a_1+3a_2) > 4a_2 > 4a_2 - 4a_1$. Therefore $a_2 = 11a_1$, and the general solution is this case is $\boxed{(a,11a,19a,29a)}$. If $k=4$, with the same reasoning we have $6a_1+2a_2 = 4a_2 - 4a_1 \Rightarrow a_2 = 5a_1$, and the general solution in this case is $\boxed{(a,5a,7a,11a)}$. We claim there are no solutions with $k=3$. Indeed, from $5a_1 + a_2 \mid 4a_2 - 4a_1$ we get $5a_1 + a_2 \leq 4a_2 - 4a_1 \iff a_2 \geq 3a_1$, and now since $a_3 = \frac{3a_1+a_2}{2}$, we have $a_3 \leq a_2$, a contradiction. The solution is complete.

My way of looking at it (which is probably equivalent to some of the above): As in -[]-'s solution, at least two of the sums must be strictly larger than $s_A/2$, so $n_A \leq 4$. For equality to hold, we must have $a_1+a_4=a_2+a_3=s_A/2.$ Writing $s_A=2n$, this means our set has the form \[\{a,b,n-b,n-a\}.\] for some $a<b<n/2$. We must have $a+b=2n/k$ and $a+n-b=2n/l$. for some positive integers $k$ and $l$, each of which must be at least $3$ by our assumption $a<b<n/2$. Our assumption $b<n/2$ also implies $k<l$. Adding, we have \[n+2a=n(\frac{2}{k}+\frac{2}{l}) >n.\] So $1/k+1/l>1/2$. The only solutions to this are $k=3, l=4$ and $k=3, l=5$. These lead to the two solutions given above.

lchserious wrote: The answers are $(1,5,7,11),(1,11,19,29),(2,10,14,22)$ and their multiples. In fact, by applying several inequalities, we know that the answer has the form $(a,b,d-b,d-a)$ where $\frac{d}{2}>b>a,(a,b)=1$. To achieve the maximum value 4, we further need $d-(b-a)|2d$ and $a+b|2d$, but since $d-(b-a)>\frac{d}{2}$, we must have $3(a-b)=d$. All we need to do is to solve $a+b|3(b-a)$, which is pretty easy. In fact, $(2,10,14,22)=2(1,5,7,11)$, so you didn't need to mention it. Very easy problem, despite being problem 1...

All the possible divisibilities are $a_1+a_2|a_3+a_4, a_1+a_3|a_2+a_4, a_1+a_4|a_2+a_3$, $a_2+a_3|a_1+a_4, a_2+a_4|a_1+a_3$, and $a_3+a_4|a_1+a_2$. Now, assume without loss of generality $a_1<a_2<a_3<a_4$. Then the cases $a_2+a_4|a_1+a_3$ and $a_3+a_4|a_1+a_2$ are easily got rid of. And let us try to make $n_A=4$. We have $a_1+a_4|a_2+a_3, a_2+a_3|a_1+a_4\Longrightarrow a_1+a_4=a_2+a_3$. Now, we still have $a_1+a_3|a_2+a_4$. Let $s=\frac{a_2+a_4}{a_1+a_3}$. Then rearranging gives $(s-1)a_3+(s+1)a_1=2a_2$ and so, $(s-1)a_3<2a_2<2a_3\Longrightarrow (s-1)a_3=a_3$ and hence, $s=2\Longrightarrow 2(a_1+a_3)=(a_2+a_4)$. Indeed, if $a_1=a, a_2=a+\alpha, a_3=a+\alpha+\beta, a_4=a+\alpha+\beta+\gamma$. Then by the conditions $a_2+a_3=a_1+a_4, 2(a_1+a_3)=(a_2+a_4)$, we have $\gamma=\alpha=2a+\beta$. So, $A=\{a, 3a+\beta, 3a+2\beta, 5a+3\beta\}$. Now, consider the condition $a_1+a_2|a_3+a_4$. This gives $4a+\beta|8a+5\beta\Longrightarrow 4a+\beta|3\beta$. Let $t=\frac{3\beta}{4a+\beta}$. Rearranging gives $(3-t)\beta=4a>0\Longrightarrow t=1, 2$. If $t=1$, then $\beta=2a\Longrightarrow \boxed{A=\{a, 5a, 7a, 11a\}}$ and if $t=2$, $\beta=8a\Longrightarrow \boxed{A=\{a, 11a, 19a, 29a\}}$

I worried that they'll cut the point for the not-stating the permutations $(a,5a,7a,11a)$ $(5a,7a,11a,a)$ and so on. I hope not. Easy and nice problem!

Raja Oktovin wrote: I worried that they'll cut the point for the not-stating the permutations $(a,5a,7a,11a)$ $(5a,7a,11a,a)$ and so on. I hope not. Easy and nice problem! Don't worry, the problem asks for the set and permutation of the elements of the set is the same set.

Icherious, your last divisibility is wrong. We have $a+b|6(a-b)$. Also how did you conclude $d=3(b-a)$?

But $\{1,3,2k-3,2k-1\}$ is also a solution since still $n=4$.

There are total $6$ possible pairs.Let wlog $a_1<a_2<a_3<a_4$. Since $a_2+a_4|s_A=a_1+a_2+a_3+a_4\implies a_2+a_4|a_1+a_3<a_2+a_4$, same for $a_3+a_4$, $n_{A}=4$. Now $a_1+a_4|a_2+a_3\implies a_1+a_4\le a_2+a_3$, if $a_2+a_3<s<2(a_2+a_3)$, $s$ is not a multiple of $a_2+a_3$. So $a_1+a_4=a_2+a_3=k,s=2k$. This shows that the set is $\{a,b,d-a,d-b\}$ for some $a,b$. We have $d-b+a|d+b-a\implies d-b+a|2d$ and $a+b|2d-a-b\implies a+b|2d$. From the first, if $d+a-b|d-a+b$ we easily conclude from $\frac{a_2+a_4}{a_1+a_3}=l,l=2\implies d=3(b-a)$ or $d=a+b$ gives the solution $a+b=2(a-b)\implies b=3a\implies a=1,b=3$. Then $a+b=4|2d\implies d=2k$ is a solution. Else $d=3(b-a)\implies a+b|6(b-a)\implies a+b|12a\implies 1<a+b|12\implies a+b=3,6,12$. We have three cases $a=1,b=2,3|d$ giving $\{1,2,3k-2,3k-1\}$. Or, $a=1,b=5$ as $\gcd(a,b)=1$. Then $6|2d\implies d=3k$ giving $\{1,5,3k-5,3k-1\}$. Else $a=1,b=11,6|d$ gives $\{1,11,6k-11,6k-1\}$. They are all solutions.ion. Else $d=3(b-a)\implies a+b|6(b-a)\implies a+b|12a\implies 1<a+b|12\implies a+b=3,6,12$. We have three cases $a=1,b=2,3|d$ giving $\{1,2,3k-2,3k-1\}$. Or, $a=1,b=5$ as $\gcd(a,b)=1$. Then $6|2d\implies d=3k$ giving $\{1,5,3k-5,3k-1\}$. Else $a=1,b=11,6|d$ gives $\{1,11,6k-11,6k-1\}$. They are all solutions.

Sorry! I forgot the condition $a_1+a_3|a_2+a_4$, employing this we find the solutions as before.

mathmdmb wrote: Icherious, your last divisibility is wrong. We have $a+b|6(a-b)$. Also how did you conclude $d=3(b-a)$? We have $d-(b-a)|2d$ and $\frac{2d}{2}=d>d-(b-a)>\frac{d}{2}=\frac{2d}{4}$. Therefore $d-(b-a)=\frac{2d}{3}$ and $3(b-a)=d$. And you are right about $a+b|3(b-a)$, it is a typo I also reverse some of the order of $a,b$ as well...

The solution to this problem is related to the root systems of the Coxeter groups F4 and H4. The size of Weyl chambers for root systems of these groups are (1,5,7,11) and (1,11,19,29) and the sum of the four number is the Weyl number. There is also a correspondence to the platonic solids. The sums (1+5, 1+7, 1+11) correspond to the number of faces, vertices and edges of the cube, while (1+11,1+19,1+29) correspond in the same way to the dodecahedron. The octohedron and icosohedron give the same solutions but the tetraedron would give (1,3,3,5) from the coxeter group D4, which fails to fit the original problem only because two numbers are the same due to the tetrahedron being self-dual. These relationships are not coincidental. They are understood in the context of Arnold's trinities. I guess that this is where the problem came from. If there was a solution to the IMO problem that made use of these correspondences it might be worth a special prize, but sadly I can't see how to get to the platonic solids directly from the divisibility conditions

The answer is $n_A=4$ with quadruples $(k, 5k, 7k, 11k)$ and $(k, 11k, 19k, 29k)$ serving as the only construction. First, note that $n_A < 5$ because $a_3+a_4 > a_1+a_2$ and $a_2+a_4 > a_1+a_3$. The rest is just a long computation. Note that for $n_A=4$ we must have $a_2+a_3 = a_1+a-4$, so $$a_1+a_3 \mid 2a_2+a_3-a_1 - (a_1+a_3) = 2(a_2-a_1).$$As $a_1+a_3 > a_2-a_1$, this implies $a_3 = 2a_2-3a_1$. Similarly, $a_1+a_2 \mid 5a_2 - 7a_1$, so it follows that $5a_2-7a_1 = k(a_1+a_2)$ where $k \in \{1, 2, 3, 4\}$. Checking all the cases yields the above two valid quadruples.

The maximal value of $n_A$ is 4, as 5 and 6 are unattainable as \[\frac{a_1+a_2+a_3+a_4}{2} < a_2+a_4, a_3+a_4 < a_1+a_2+a_3+a_4,\] where we assume WLOG $a_1<a_2<a_3<a_4$. To attain 4, we notice we must have \[a_1+a_4 \mid a_2+a_3, \quad a_2+a_3 \mid a_1+a_4 \implies a_1+a_4 = a_2+a_3.\] Hence our set can be rewritten as $\{a,b,x-b,x-a\}$, where $x<2b$, and we also require \[x+a-b \mid 2x-2(x+a-b) = 2(b-a) \implies x+a-b = 2(b-a) \implies x = 3b-3a.\]\[a+b \mid 2x = 6(b-a) \implies a+b \mid 12a \implies 12a = (a+b), \ldots, 5(a+b).\] Testing all 5 cases, we find the only solutions $\boxed{(a,5a,7a,11a),~(a,11a,19a,29a) \quad a \in \mathbb{N}}$. $\blacksquare$

The answer is \(n_A = 4\), achieved by the sets \(A = \{t, 11t, 19t, 29t\}\) and \(A = \{t, 5t, 7t, 11t\}\) for any positive integer \(t\). One can check that they work. Without loss of generality, \(a_1 < a_2 < a_3 < a_4\). Claim 1. We have \(n_A \leq 4\). Proof. If we have two positive integers \(a\) and \(b\) that divide their sum, then they are equal to each other. Similarly, if \(n_A > 4\), then at least two of \((a_1 + a_2, a_3 + a_4)\), \((a_1 + a_3, a_2 + a_4)\) and \((a_1 + a_4, a_2 + a_3)\) have equal sums. However, \(a_1 + a_2 < a_3 + a_4\) and \(a_1 + a_3 < a_2 + a_4\), so we have a contradiction. \(\square\) In particular, if \(n_A = 4\), then \(a_1 + a_4 = a_2 + a_3\). Claim 2. We achieve \(n_A = 4\) with only the sets \(A = \{t, 11t, 19t, 29t\}\) and \(A = \{t, 5t, 7t, 11t\}\) for any positive integer \(t\). Proof We let \(a_1 = x\), \(a_2 = x + y\), \(a_3 = x + y + z\) and \(a_4 = x + y + z + y\) to homogenize the condition \(a_1 + a_4 = a_2 + a_3\). In order to achieve \(n_A = 4\), we need: \[2x + y = a_1 + a_2 \mid s_A = 4x + 4y + 2z\]\[2x + y + z = a_1 + a_3 \mid s_A = 4x + 4y + 2z\]This is equivalent to \(2x + y \mid 2y + 2z\) and \(2x + y + z \mid 2y\). Notice how the second condition implies \(2x + y + z = 2y\), because \(2x + y + z > y\), which is the second largest possible divisor of \(2y\). We get \(2x + z = y\). Substituting that into the first condition, we obtain \[4x + z \mid 4x + 4z \implies 4x + z \mid 3z.\]using the same argument as before, we have \(4x + z > z\), so either \(4x + z = \frac{3}{2}z\) or \(4x + z = 3z\). In other words, \(8x = z\) or \(2x = z\). This implies \(10x = y\) or \(4x = y\). We now reverse the substitution at the beginning of the proof to get \[(a_1, a_2, a_3, a_4) \in \{(t, 11t, 19t, 29t), (t, 5t, 7t, 11t)\}\]completing the solution. \(\square\)

Without loss of generality let $a_1 < a_2 < a_3 < a_4$. Note then that clearly, \begin{align*} a_3 + a_4 < s_A < 2(a_3 + a_4) \end{align*}and hence $a_3 + a_4 \nmid s_A$. Similarly as, \begin{align*} a_1 + a_3 &\leq a_2 + a_4 \iff (a_3 - a_4) \leq (a_2 - a_1) \end{align*}we cannot have $a_2 + a_4 \mid s_A$. This proves the bound of $n_A = 4$. Now it remains to find all sets $A$ of four distinct positive integers that achieve $n_A = 4$. Begin by noting we must have, $a_2 + a_3 \equiv 0 \pmod{a_1 + a_4}$ $a_1 + a_4 \equiv 0 \pmod{a_2 + a_3}$ $a_3 + a_4 \equiv 0 \pmod{a_1 + a_2}$ $a_2 + a_4 \equiv 0 \pmod{a_1 + a_3}$ It immediately follows from the first two bullets that $a_1 + a_4 = a_2 + a_3$. From the remaining bullets we substitute $a_4 = a_2 + a_3 - a_1$ to find, \begin{align*} a_3 + a_4 \equiv a_2 + 2a_3 - a_1 &\equiv 0 \pmod{a_1 + a_2} \iff a_1 + a_2 \mid a_2 + 2a_3 - a_1\\ a_2 + a_4 \equiv 2a_2 + a_3 - a_1 &\equiv 0 \pmod{a_1 + a_3} \iff a_1 + a_3 \mid 2a_2 + a_2 - a_1 \end{align*}The relations then become, \begin{align*} a_1 + a_2 \mid a_2 + 2a_3 - a_1 \iff a_1 + a_2 \mid 2( a_3 - a_1)\\ a_1 + a_3 \mid 2a_2 + a_3 - a_1 \iff a_1 + a_3 \mid 2(a_2 - a_1) \end{align*}Thus let, \begin{align*} m(a_1 + a_2) = 2(a_3 - a_1)\\ n(a_1 + a_3) = 2(a_2 - a_1) \end{align*}This rearranges to, \begin{align*} (m + 2)a_1 = 2a_3 - ma_2\\ (n + 2)a_1 = 2a_2 - na_3 \end{align*}Note that the last equation forces $n = 1$ as for $n > 2$ we find $2a_2 - 2a_3 < 0$. Thus we have, \begin{align*} (m + 2)a_1 &= 2a_3 - ma_2\\ 3a_1 &= 2a_2 - a_3 \end{align*}Now we have, \begin{align*} (m + 2)a_1 &= 2(a_3 - 2a_2) - (m - 4)a_2\\ (m + 8)a_1 &= (4 - m)a_2\\ \frac{a_1}{a_2} &= \frac{4 - m}{m + 8} \end{align*}Note that clearly $m < 4$. Thus we find that, \begin{align*} \frac{a_1}{a_2} \in \left\{ \frac{1}{3}, \frac{1}{5}, \frac{1}{11} \right\} \end{align*}We're now at the home-stretch. We take cases on $\frac{a_1}{a_2}$. Case 1: If $a_2 = 3a_1$, then $a_3 = 3a_1$ contradiction as then $a_2 = a_3$. Case 2: If $a_2 = 5a_1$, then $a_3 = 7a_1$ and $a_4 = 1 1a_1$. This can be checked to always work. Case 3: If $a_2 = 1 1a_1$, then $a_3 = 19a_1$ and hence $a_4 = 29a_1$, and this always works as well. Our solution set is then for $a \in \mathbb{Z}^+$: \begin{align*} \boxed{\left\{ (a_1, 5a_1, 7a_1, 11a_1), (a_1, 11a_1, 19a_1, 29a_1) \right\}} \end{align*}

WLOG let $a_1<a_2<a_3<a_4$. Since $a_1+a_2<a_3+a_4$ and $a_1+a_3<a_2+a_4$, $n_A\le4$, with equality only possible when $a_1+a_4=a_2+a_3$. Let the numbers be $\{a, b, n-b, n-a\}$ for $0<a<b<\frac{n}2$. We want $(n-(b-a))|(2n)$. $n-(b-a)\ge \frac{2n}{2}$ and $n-(b-a)\le \frac{2n}{4}$ are clearly absurd, so $n-(b-a)=\frac{2n}3$ or $b-a=\frac{n}{3}$. We also want $(a+b)|(2n)$. As $b-a=\frac{n}{3}$, $\frac{2n}{3}>a+b>\frac{2n}{6}$ so $a+b=\frac{2n}4=\frac{n}{2}$ or $a+b=\frac{2n}{5}$. The first case gives $\{k, 5k, 7k, 11k\}$ and the second case gives $\{k, 11k, 19k, 29k\}$.

The answers are $a_1:a_2:a_3:a_4 = 1:5:7:11$ and $a_1:a_2:a_3:a_4 = 1:11:19:29$, both of which clearly work. Sort $a$. Consider the pairs $a_1 + a_2, a_3+ a_4$ and $a_1 + a_3, a_2 + a_4$. One of the elements in each pair is clearly larger than the other, so at most one can divide into the total sum. Then if we want $n_a = 4$, we need $a_1 + a_4 = a_2 + a_3$. Thus let $a_1 = x, a_2 = y, a_3 = z$, we then require $x + y \mid 2(y + z), x + z \mid 2(y + z)$. Let $x + y = d, x + z = e, y + z = f$, then we require $d \mid 2f, e \mid 2f$, and $d,e,f$ satisfy the triangle inequality to make $x,y,z$ positive. By size then, we clearly need $e > \frac f2$, so $e = \frac{2f}{3}$. Since $d,e$ are distinct, we can either have $d = \frac{2f}{4}, d = \frac{2f}{5}$. In the first case, we get $d:e:f = 6:8:12, x:y:z = 1:5:7$, so $a_1:a_2:a_3 = 1:5:7$, since $a_1 + a_4 = a_2 + a_3$ gives $a_1:a_2:a_3:a_4 = 1:5:7:11$. In the second case, we get $d:e:f = 12:18:20, x:y:z = 1:11:19$, so $a_1:a_2:a_3:1:11:19$, since $a_1 + a_4 = a_2 + a_3$, gives $a_1:a_2:a_3:a_4 = 1:11:19:29$.

WLOG assume that $a_1>a_2>a_3>a_4$. We have $n_A\leq 4$ since $a_1+a_2\nmid a_3+a_4$ and $a_1+a_3\nmid a_2+a_4$. Notice that $n_A=4$ is achievable: $5+1\mid 11+7$, $7+1\mid 11+5$, $7+5\mid 11+1$, $11+1\mid 7+5$. Now, we need to find all $A$ such that $n_A=4$. In that case we must have $$a_1+a_4=a_3+a_2, \ a_3+a_4\mid a_1+a_2, \ a_2+a_4\mid a_1+a_3 \qquad (1)$$Take $a_4=a$, $a_3=a+b$, $a_2=a+b+c$, $a_1=a+b+c+d$. Then $$a_1+a_4=a_3+a_2\implies b=d.$$Now, $(1)$ implies that (replacing $d$): \begin{align*} & 2a+b \mid 2a+3b+2c \Rightarrow 2a+b\mid 2b+2c \qquad (2) \\ & 2a+b+c \mid 2a+3b+c \Rightarrow 2a+b+c\mid 2b \qquad (3)\end{align*}$(3)$ implies that $2a+b+c=2b$, since $2(2a+b+c)>2b$. So $c=b-2a$ and so $(2)$ implies that $2a+b\mid 4b-4a$. If $2a+b=4b-4a$ then $b=2a$ but then $c=0$, a contradiction. Thus either $2(2a+b)=4b-4a$ or $3(2a+b)=4b-4a$. In the first case $b=4a$, $c=2a$ and in the second $b=10a$, $c=8a$. Returning back we get $$(a_1,a_2,a_3,a_4)\in \{(29a, 19a, 11a, a), (11a, 7a, 5a, a)\}.$$Thus the answer is $\{29k,19k,11k,k\}$ for any $k\in \mathbb N$ and $\{11k,7k,5k,k\}$ for any $k\in \mathbb N$. These clearly work.

Another old solution, which I am posting for storage. We claim that the maximum of $n_A$ is $4$. It is easy to see that this $n_A\leq 4$ since always $a_2+a_4>a_1+a_3$ and $a_3+a_4>a_1+a_2$ which contradicts two of the six possible divisibilities. Now, we find all sets for which $n_A=4$ is achievable. Note that we must have, $a_1+a_4\leq a_2+a_3$ and $a_2+a_3 \leq a_1+a_4$ which implies that $a_4=a_2+a_3-a_1$. Further, \[a_1+a_3 \mid a_2+a_4 \implies a_1+a_3 \mid 2a_2+a_3-a_1 \implies a_1+a_3 \mid 2a_2-2a_1\]Now, $2a_2-2a_1 < 2a_3-2a_1<2a_3+2a_1=2(a_1+a_3)$ Thus, $2a_2-2a_1=0 \implies a_1=a_2$ which is impossible or $2a_2-2a_2=a_1+a_3$ which gives us that $a_3=2a_2-3a_1$. Now note that this means $a_1+a_2\mid a_3+a_4 \implies a_1+a_2 \mid 12a_2$. Thus, $a_1=\frac{(12-k)a_2}{k}$. Thus, clearly $k<12$. Further, $a_4=3a_2-4a_1 \implies 3a_2\geq \frac{(48-4k)a_2}{k}\implies k\geq 8$. Now, we check. It is easy to check that $k=8$ and $k=9$ yield no solutions. $k=10$ yields the solution set $\{r,5r,7r,11r\}$ for all positive integers $r$ and $k=11$ gives us $\{r,11r,19r,29r\}$ for all positive integers $r$. Thus, these are the only solutions and we are done.

First let us relable the variables as $a,b,c,d$. WLOG let $a<b<c<d$. Note that for any pair, say $a+c$ to divide $a+b+c+d$ it is sufficient that they divide $a+b+c+d-a-c=b+d$. Note now that $a+b<c+d$ so at best this breakup contributes one division via $a+b | c+d$, similarly we have $a+c<b+d$ so at best one with $a+c|b+d$, then we can have $a+d=b+c$ contributing two divisions $a+d|b+c,b+c|a+d$. Thus $n_A\leq 4$. Now we will derive all tuples satisfying $n_A =4$. Note that $a+d=b+c \implies d-c=b-a=x$. Rewriting we get $a,a+x,c,c+x$ and we want to satisfy $2a+x|2c+x, a+c|a+c+2x$. The second divisibility gives ups $a+c|2x$. Note that if $a+c<2x$, then $a+c \leq x \implies x \geq a+a+(c-a)>2a+x$ which is impossible so $a+c=2x$. From the former divisibility, since $2a+x<2c+x$ we have $2a+x|2c+x-2a-x=2(c-a)=2(2x-2a)=4(x-a)$. Thus $2a+x|4(x-a) \implies 2|x$, let $x=2k$ we have $a+k|2(2k-a)$. Now, if $a+k =4k-2a$ we will have $3a=3k \implies a=k \implies k+c=4k \implies c=3k$. However, then $b=3k=c$ is a contradiction. Thus we have $a+k<4k-2a \implies a+k|3k-3a$. If equality holds in this case we have $2a=k$, letting $k=2t$ would give us the set of solutions $(t,5t,7t,11t)$. If equality doesn't hold in this case then $a+k|2k-4a$. If equality holds here then $5a=k$, letting $k=5t$ gives us the set of solutions $(t, 11t, 19t, 29t)$. If equality doesn't hold then $a+k|k-5a$ which is absurd because this would require $6a\leq 0$ while $a>0$. Thus the set of numbers that work are $(t,5t,7t,11t),(t,11t,19t, 29t)$ and all their permutations for every positive integer $t$.

Don't know why, but this feels like an easy A1 despite this being a long solution The sets $\{a_1,a_2,a_3,a_4\}$ that satisfy the property that $p_A$ is maximal are $$\{x,5x,7x,11x\}$$and $$\{y,11y,19y,29y\}$$for any positive integers $x$ and $y$, and permutations of these. Notice that these two families of sets both yield $p_A=4$. This is maximal as $(i,j)=(2,4)$ and $(i,j)=(3,4)$ are clearly both invalid (where we assume that $a_1<a_2<a_3<a_4$). This only leaves $\dbinom{4}{2}-2=4$ pairs $(i,j)$ remaining, so this is maximal and attainable. We prove that these are the only solutions with maximal $p_A$. Suppose otherwise, and by WLOG that $$a_1<a_2<a_3<a_4.$$Notice that $$a_1+a_2+a_3+a_4\equiv 0\pmod {a_1+a_4}\implies a_2+a_3\equiv 0\pmod {a_1+a_4}.$$Similarly, we have $$a_1+a_4\equiv 0\pmod {a_2+a_3}.$$It is clear that $$a_1+a_4=a_2+a_3\implies a_4=a_2+a_3-a_1.$$Note that $$2<\frac{a_1+a_2+a_3+a_4}{a_1+a_3}=2\cdot\frac{a_2+a_3}{a_1+a_3}<2\cdot \frac{2a_3}{a_1+a_3}<\frac{4a_3}{a_3}=4,$$so $$\frac{a_1+a_2+a_3+a_4}{a_1+a_3}=3$$$$\implies 2a_1+2a_3=a_2+(a_2+a_3-a_1)$$$$\implies 2a_1+2a_3=2a_2+a_3-a_1$$$$\implies 3a_1+a_3=2a_2$$$$\implies a_3=2a_2-3a_1.$$Now, we know that $$a_1+a_2+a_3+a_4$$$$=a_1+a_2+2a_2-3a_1+a_2+(2a_2-3a_1)-a_1$$$$=6a_2-6a_1.$$Additionally, we have $$2<\frac{6a_2-6a_1}{a_1+a_2}<6.$$We split this into casework: First, we have $$\frac{6a_2-6a_1}{a_1+a_2}=3\implies 6a_2-6a_1=3a_1+3a_2\implies a_2=3a_1.$$However, this would give $$a_3=6a_1-3a_1=3a_1,$$but we need $a_2<a_3$. There are no solutions here. Second, we have $$\frac{6a_2-6a_1}{a_1+a_2}=4\implies 6a_2-6a_1=4a_1+4a_2\implies a_2=5a_1.$$This gives the $$\{x,5x,7x,11x\}$$solution discussed earlier (with the $a_1$ being $x$). Finally, we have $$\frac{6a_2-6a_1}{a_1+a_2}=5\implies 6a_2-6a_1=5a_1+5a_2\implies a_2=11a_1.$$This gives the $$\{y,11y,19y,29y\}$$solution discussed earlier (with the $a_2$ being $y$). This concludes the problem.

Why is this 10 MOHS :skull: The largest possible value of $n_{A}$ is $4.$ First, we will show that no higher $n_{A}$ is possible. Assume for the sake of contradiction that $n_{A} \ge 5.$ Then, there would be at least two sets of permutations $\{i, j, k, \ell\}$ of $\{1, 2, 3, 4\}$ where $a_{i}+a_{j}, a_{k}+a_{\ell} \mid s_{A}.$ Notice that $a_{k}+a_{\ell}, a_{i}+a_{j} \neq s_{A},$ so $$a_{k}+a_{\ell}, a_{i}+a_{j} \le \frac{s_{A}}{2}.$$But, $$a_{i}+a_{j}+a_{k}+a_{\ell} = \frac{s_{A}}{2},$$so $a_{i}+a_{j} = a_{k}+a_{\ell} = \frac{s_{A}}{2}.$ So, we have that $$a_{1}+a_{i} = \frac{s_{A}}{2},$$and $$a_{1}+a_{j} = \frac{s_{A}}{2},$$where $i \neq a_{j}.$ But, then this contradicts the fact that the numbers in $A$ are distinct. Thus, $n_{A} \le 4.$ Next, we find when $n_{A} = 4.$ We have that $$a_{1}+a_{i} = a_{j}+a_{k} = \frac{s_{A}}{2},$$where $i \neq j \neq k,$ and $i,j,k \in \{2,3,4\}.$ Since the order of the numbers does not matter, we can assume WLOG that $a_{1}+a_{2} = \frac{s_{A}}{2}.$ Now, we have two other subsets of $A$ with cardinality $2$ that have their sum a divisor of $s_{A}.$ We know that they must share an element, since otherwise we will get two of the "permutations" from the proof that $n_{A} \le 4$ earlier. Assume WLOG that this is common element is $a_{1}.$ Then, $$a_{1}+a_{3} = \frac{s_{A}}{a},$$and $$a_{1}+a_{4} = \frac{s_{A}}{b}.$$So, $$a_{4}-a_{3} = \frac{s_{A}}{b}-\frac{s_{A}}{a},$$and hence $$a_{4} = s_{A}\left(\frac{1}{4}+\frac{1}{2b}-\frac{1}{2a} \right)$$and $$a_{4} = s_{A}\left(\frac{1}{4}-\frac{1}{2b}+\frac{1}{2a} \right).$$Now, $$a_{1} = s_{A}\left(-\frac{1}{4}+\frac{1}{2b}+\frac{1}{2a} \right),$$and $$a_{2} = s_{A}\left(\frac{3}{4}-\frac{1}{2b}-\frac{1}{2a}\right).$$So, we need to have $$\frac{3}{2} > \frac{1}{a}+\frac{1}{b} > \frac{1}{2},$$and $$\frac{1}{a}+\frac{1}{2} > \frac{1}{b} > \frac{1}{a} - \frac{1}{2}.$$Notice that $a,b \ge 3,$ since $a,b \ge 2,$ and $a,b \neq 2.$ So, the last inequality chain is redundant. In fact, the only meaningful inequality is $$\frac{1}{a}+\frac{1}{b} > \frac{1}{2}.$$Assume WLOG that $a \ge b$ (this will only swap $a_{3}$ and $a_{4},$ which have been symmetric until now). Then, $a < 4,$ so $a = 3.$ So, $b = 3, 4, 5.$ If $(a,b) = \left(3, 3 \right),$ then $a_{3}, a_{4} = \frac{s_{A}}{4},$ and so $a_{1} = \frac{s_{A}}{12},$ and thus $a_{2}=\frac{5s_{A}}{12}.$ So, $$A = \{n, 3n, 3n, 5n\},$$where $n$ is a positive integer. If $(a,b) = \left(3, 4\right),$ we get that $A = \{n, 11n, 5n, 7n\}.$ If $(a,b) = \left(3,5\right),$ we find that $A = \{n, 29n, 19n, 11n\}.$ So, we have completed the proof.

wow, three posts at around the same time?!?! :O Clearly if we have at least $5$ pairs there will be two "conjugate" pairs of pairs that each divide $s,$ implying that some two numbers are equal, contradiction. Thus $n_A =4.$ WLOG assume that $a_1 \leq a_2 \leq a_3 \leq a_4.$ Then $a_i+a_j$ divides $s$ if and only if $a_i+a_j \leq \frac{s}{2}.$ Thus $a_1+a_4, a_2+a_3, a_1+a_2, a_1+a_3$ are the pairs that divide $s.$ From the first two it follows that $a_1+a_4=a_2+a_3,$ thus $s=2(a_2+a_3).$ The latter two pairs thus imply $$\frac{2(a_3-a_1)}{a_1+a_2}, \frac{2(a_2-a_1)}{a_1+a_3} \in \mathbb Z^+.$$But, observe that their product is $$\frac{4(a_3-a_1)(a_2-a_1)}{(a_1+a_2)(a_1+a_3)} < 4.$$Therefore this product can only take the values $1, 2, 3.$ We now proceed with casework. $\text{Case 1: Product is 1.}$ Then the two parts of the product are each equal to $1.$ Hence $2(a_3-a_1)=a_1+a_2, 2a_2-2a_1=a_1+a_3 \implies 2a_3=3a_1+a_2, 2a_2=3a_1+a_3.$ But this yields $a_2=a_3,$ contradiction. $\text{Case 2: Product is 2.}$ Then the parts of the product are $1, 2.$ The only way this can happen is if $a_3-a_1=a_1+a_2, 2a_2-2a_1=a_1+a_3.$ These yield $a_3=2a_1+a_2, 2a_2=3a_1+a_3.$ Thus $a_2=5a_1, a_3=7a_1, \implies a_4=11a_1.$ Hence we have the solution $(k, 5k, 7k, 11k)$ for positive integers $k.$ $\text{Case 3: Product is 3.}$ Then the parts of the product are $1, 3.$ Thus $2a_3=5a_1+3a_2, 2a_2=3a_1+a_3.$ This gives $a_2=11a_1, a_3=19a_1, a_4=29a_1.$ Thus we have the solution $(k, 11k, 19k, 29k)$ for positive integers $k.$ To summarize, the solutions are $(k, 5k, 7k, 11k), (k, 11k, 19k, 29k)$ for positive integers $k.$

Mr.Sharkman wrote: Why is this 10 MOHS :skull: The largest possible value of $n_{A}$ is $4.$ First, we will show that no higher $n_{A}$ is possible. Assume for the sake of contradiction that $n_{A} \ge 5.$ Then, there would be at least two sets of permutations $\{i, j, k, \ell\}$ of $\{1, 2, 3, 4\}$ where $a_{i}+a_{j}, a_{k}+a_{\ell} \mid s_{A}.$ Notice that $a_{k}+a_{\ell}, a_{i}+a_{j} \neq s_{A},$ so $$a_{k}+a_{\ell}, a_{i}+a_{j} \le \frac{s_{A}}{2}.$$But, $$a_{i}+a_{j}+a_{k}+a_{\ell} = \frac{s_{A}}{2},$$so $a_{i}+a_{j} = a_{k}+a_{\ell} = \frac{s_{A}}{2}.$ So, we have that $$a_{1}+a_{i} = \frac{s_{A}}{2},$$and $$a_{1}+a_{j} = \frac{s_{A}}{2},$$where $i \neq a_{j}.$ But, then this contradicts the fact that the numbers in $A$ are distinct. Thus, $n_{A} \le 4.$ Next, we find when $n_{A} = 4.$ We have that $$a_{1}+a_{i} = a_{j}+a_{k} = \frac{s_{A}}{2},$$where $i \neq j \neq k,$ and $i,j,k \in \{2,3,4\}.$ Since the order of the numbers does not matter, we can assume WLOG that $a_{1}+a_{2} = \frac{s_{A}}{2}.$ Now, we have two other subsets of $A$ with cardinality $2$ that have their sum a divisor of $s_{A}.$ We know that they must share an element, since otherwise we will get two of the "permutations" from the proof that $n_{A} \le 4$ earlier. Assume WLOG that this is common element is $a_{1}.$ Then, $$a_{1}+a_{3} = \frac{s_{A}}{a},$$and $$a_{1}+a_{4} = \frac{s_{A}}{b}.$$So, $$a_{4}-a_{3} = \frac{s_{A}}{b}-\frac{s_{A}}{a},$$and hence $$a_{4} = s_{A}\left(\frac{1}{4}+\frac{1}{2b}-\frac{1}{2a} \right)$$and $$a_{4} = s_{A}\left(\frac{1}{4}-\frac{1}{2b}+\frac{1}{2a} \right).$$Now, $$a_{1} = s_{A}\left(-\frac{1}{4}+\frac{1}{2b}+\frac{1}{2a} \right),$$and $$a_{2} = s_{A}\left(\frac{3}{4}-\frac{1}{2b}-\frac{1}{2a}\right).$$So, we need to have $$\frac{3}{2} > \frac{1}{a}+\frac{1}{b} > \frac{1}{2},$$and $$\frac{1}{a}+\frac{1}{2} > \frac{1}{b} > \frac{1}{a} - \frac{1}{2}.$$Notice that $a,b \ge 3,$ since $a,b \ge 2,$ and $a,b \neq 2.$ So, the last inequality chain is redundant. In fact, the only meaningful inequality is $$\frac{1}{a}+\frac{1}{b} > \frac{1}{2}.$$Assume WLOG that $a \ge b$ (this will only swap $a_{3}$ and $a_{4},$ which have been symmetric until now). Then, $a < 4,$ so $a = 3.$ So, $b = 3, 4, 5.$ If $(a,b) = \left(3, 3 \right),$ then $a_{3}, a_{4} = \frac{s_{A}}{4},$ and so $a_{1} = \frac{s_{A}}{12},$ and thus $a_{2}=\frac{5s_{A}}{12}.$ So, $$A = \{n, 3n, 3n, 5n\},$$where $n$ is a positive integer. If $(a,b) = \left(3, 4\right),$ we get that $A = \{n, 11n, 5n, 7n\}.$ If $(a,b) = \left(3,5\right),$ we find that $A = \{n, 29n, 19n, 11n\}.$ So, we have completed the proof. yes this was 0 mohs where are isl mohs ratings

Bro this is IMO 2011/1 :skull: But I think that this should have been N1, and not A1

Mr.Sharkman wrote: Bro this is IMO 2011/1 :skull: But I think that this should have been N1, and not A1 oh im sobad no there is very little nt