Basically as IMO 2011 P1. For distinct indices $i,j,k,m,n$ we have $x_i + x_j + x_k \mid x_i + x_j + x_k + x_m + x_n \Leftrightarrow x_i + x_j + x_k \mid x_m + x_n$. Without loss of generality let $x_1 < x_2 < x_3 < x_4 < x_5$. Then $x_3 + x_4 + x_5 > x_1 + x_2$, $x_2 + x_4 + x_5 > x_1 + x_3$, $x_1 + x_4 + x_5 > x_2 + x_3$, $x_1 + x_3 + x_5 > x_2 + x_4$ and $x_2 + x_3 + x_5 > x_1 + x_4$. Also, $x_1 + x_2 + x_5 \leq x_3 + x_4$ and $x_2 + x_3 + x_4 \leq x_1 + x_5$ cannot simultaenously hold since otherwise adding them and cancelling gives $x_2 \leq 0$, which is false. Therefore we can have at most $4$ sums. An example with $4$ sums is with $x_1 = 1$, $x_2 = 2$, $x_3 = 3$, $x_4 = 4$ and $x_5$ be such that $6$, $7$, $8$ and $9$ divide $x_5 + 10$, so e.g. $x_5 = 6 \cdot 7 \cdot 8 \cdot 9 - 10 = 3014$. (An alternative is $x_5 = \mbox{lcm}(6,7,8,9) - 10 = 494.$)

I think the answer should be $4$. Here's a quick sketch. If we have at least $5$, then if without loss of generality the elements are sorted, i.e. $x_{1}<...<x_{5}$, then we must have $x_{5}$ in one such triple. Then of course the sum of the triple divides the sum of the other two elements, so this directly eliminates all but $(1,2,5)$ due to size issues (for example $(1,3,5)$ is not possible because $x_{5}>x_{4}$ and $x_{3}>x_{2}$ so $x_{1}+x_{3}+x_{5} > x_{2}+x_{4}$). This case is also not possible because then we have all other triples without $x_{5}$, so $x_{1}+x_{2}+x_{5} \leq x_{3} + x_{4}$ and $x_{2}+x_{3}+x_{4} \leq x_{1} + x_{5}$. Adding these we get $2x_{2} \leq 0$, a contradiction. To construct $4$ fix $x_{1} = a$,..., $x_{4} = d$ and take for example $S =100(a+b+c)(a+b+d)(a+c+d)(b+c+d)$, which is divisible by the sums of the four smallest triples. Now just take $x_{5} = e = S-a-b-c-d > d > 0$. Note: There is a lot of freedom in choosing the construction.

Notice that $x_{i_1}+x_{i_2}+x_{i_3} \mid S_A=x_{i_1}+x_{i_2}+x_{i_3}+x_{i_4}+x_{i_5}$ iff $ x_{i_1}+x_{i_2}+x_{i_3} \mid x_{i_4}+x_{i_5}$ Then WLOG, let $x_1<x_2<x_3<x_4<x_5$ then these inequality hold: \[\left\{ \begin{array}{l} {x_5} + {x_4} + {x_3} > {x_1} + {x_2}\\ {x_5} + {x_4} + {x_2} > {x_1} + {x_4}\\ {x_5} + {x_4} + {x_1} > {x_2} + {x_3}\\ {x_5} + {x_3} + {x_2} > {x_1} + {x_4}\\ {x_5} + {x_3} + {x_1} > {x_2} + {x_4} \end{array} \right.\]And also $x_5+x_2+x_1 \mid x_3+x_4$ and $x_3+x_4+x_2 \mid x_5+x_1$ can't hold at the same time Thus there are at most $T_A \le 4$ Here is the way to construct the case $T_4 =4 $: We find $x_5$ satisfies the set of congruence: \[\left\{ \begin{array}{l} {x_5} \equiv - {x_1}\pmod {{x_2} + {x_3} + {x_4}}\\ {x_5} \equiv - {x_2}\pmod {{x_1} + {x_3} + {x_4}}\\ {x_5} \equiv - {x_3}\pmod {{x_1} + {x_2} + {x_4}}\\ {x_5} \equiv - {x_4}\pmod {{x_2} + {x_3} + {x_1}} \end{array} \right.\]And when we pick $x_1=1,x_2=2,x_3=3,x_4=4$, it's possible to find $x_5$ due to CRT, and done!!

@above the numbers are distinct, so the construction doesn't work. In addition, the modulos should be coprime as well, otherwise you can't use CRT. So you should give an explicit example.

VicKmath7 wrote: @above the numbers are distinct, so the construction doesn't work. In addition, the modulos should be coprime as well, otherwise you can't use CRT We use a stronger version of CRT, which is known as the set of equation $x \equiv r_i \pmod {m_i}$ for $1 \le i \le n$ has solution if and only if $r_i \equiv r_j \pmod {gcd(m_i,m_j)}$ and it's possible now when we pick $x_1=1,x_2=2,x_3=3,x_4=4$ Also by CRT we can pick $x_5$ large arbitrarily which is able to be distinct from the other $4$

Ok, but you still have $x_2=x_3$. So you should have probably chosen $x_2=2$ as the poster in #2 did. Edit: the post was corrected afterwards. Edit2: Btw, I now remembered an application of the stronger version of CRT, so anyone interested could see ISL 2006 N7, the solution in #32.

VicKmath7 wrote: Ok, but you still have $x_2=x_3$. So you should have probably chosen $x_2=2$ as the poster in #2 did. Edit: the post was corrected afterwards. Yes sorry for the typo!!

On another note, as others have noticed, this is very similar to Problem 1 from IMO 2011. Ideally the problems should be original, but choosing a variation of a problem from the IMO itself is very far from that, not to mention that it gives an unfair advantage to those who have seen it.

I claim the answer is 4. Note that for any triple $(i,j,k)$, it is true that \[S_A > x_i+x_j+x_k\]Thus, $S_A\geq 2\cdot (x_i+x_j+x_k)$, so $S_A - x_i - x_j - x_k \geq x_i+x_j+x_k$. This implies that the triples $(1,3,5),(1,4,5),(2,3,5),(2,4,5),(3,4,5)$ fail because they all majorise their complements: \[x_1+x_3+x_5>x_2+x_4\]\[x_1+x_4+x_5>x_2+x_3\]\[x_2+x_3+x_5>x_1+x_4\]\[x_2+x_4+x_5>x_1+x_3\]\[x_3+x_4+x_5>x_1+x_2\] Then, note that we cannot have both $(1,2,5)$ and $(1,3,4)$ working because \[x_1+x_2+x_5\leq x_3 + x_4\]\[x_1+x_3+x_4\leq x_2+x_5\]Implies $x_1\leq 0$, a contradiction. Thus, we have ruled out at least 6 of the 10 possible triplets, so $T_A \leq 10-6=4$ Note that this is achievable when $x_i=i$ for $i\leq 4$, and $x_5 = 6\cdot 7 \cdot 8\cdot 9$.

This problem was proposed by MathsLion (Boris Stankovic), Bosnia and Herzegovina

Solved with mathmax12: We claim that the answer is $\boxed{4}.$ (We couldn't find out proof that 5 doesn't work) A possibility with $4$, is $x_1 = 1$, $x_2 = 2$, $x_3 = 3$, $x_4 = 4$, $x_5=494.$

We claim that the maximum value of $T_A$ is $\boxed{4}$. It is easy to verify that this is indeed attainable; take $S =\{1,2,3,4,494\}$. We now prove that $T_A\le 4$. WLOG, let us assume that $x_1<x_2<x_3<x_4<x_5$. Claim: $T_A\le 5$. Proof: Observe that if $(\ell_1,\ell_2,\ell_3,\ell_4,\ell_5)$ is some permutation of $(1,2,3,4,5)$, then \[\frac{S}{x_{\ell_1}+x_{\ell_2}+x_{\ell_3}} = \frac{x_{\ell_1}+x_{\ell_2}+x_{\ell_3}+x_{\ell_4}+{\ell_5}}{x_{\ell_1}+x_{\ell_2}+x_{\ell_3}} = 1+\frac{x_{\ell_4}+x_{\ell_5}}{x_{\ell_1}+x_{\ell_2}+x_{\ell_3}},\]is not an integer if it is not the case that $x_{\ell_4}+x_{\ell_5} \ge x_{\ell_1}+x_{\ell_2}+x_{\ell_3}$. Now, since \begin{align*} x_1 + x_2 &< x_3+x_4+x_5 \\ x_1+x_3 &< x_2+x_4+x_5 \\ x_1+x_4 &< x_2+x_3+x_5 \\ x_2+x_3&<x_1+x_4+x_5 \\ x_2+x_4 &< x_1+x_3+x_5,\end{align*}it is impossible that any of the following are integers: \[\frac{S}{x_3+x_4+x_5}, \frac{S}{x_2+x_4+x_5}, \frac{S}{x_2+x_3+x_5}, \frac{S}{x_1+x_4+x_4}, \frac{S}{x_1+x_3+x_5}.\]This proves our claim. Claim: $T_A\ne 5$. Proof: For the sake of contradiction, assume that $T_A = 5$. Then, we must have \begin{align*} x_1+x_5&\ge x_2+x_3+x_4 \\ x_2+x_5 &\ge x_1+x_3+x_4 \\ x_3+x_4 &\ge x_1+x_2+x_5 \\ x_3+x_5 &\ge x_1+x_2+x_4 \\ x_4+x_5 &\ge x_1+x_2+x_3.\end{align*}Multiplying the second inequality by $-1$, flipping the sign, and adding it to the first, we obtain $x_5\ge x_3+x_4$. But then, by the third inequality, we must have $x_5\ge x_3+x_4 \ge x_1+x_2+x_5$, contradiction. Hence $T\ne 5$, and we are done. $\blacksquare$