Are you sure this was also at shortlist since it seems those problems for Tajikistan weren't at shortlist. I have the shortlist and these are not there could you tell me where you are getting the problems.

https://pregatirematematicaolimpiadejuniori.wordpress.com/2020/09/12/shl2019/ I guess it is from there.

Okay. Maybe something happened since on the version I have there wasn't these problems. That is why I asked.

I think this works.

I assume that it's four distinct numbers. Let the $100$ numbers be $a_1 \ge a_2 \ge ...\ge a_{100}$. Consider the four numbers: $a_{100}, a_x ,a_n,a_1$. $a_1$ is the largest, and for each $n$, there is at most one $x=x_n$ such that $a_1=a_n+a_{x_n}+a_{100}$ (If it exists). For every $n=2,3,..,99$, we choose $x \neq x_n$, and so by the condition in the question, $a_{100}|a_n$ and $a_{100}|a_{1}$. So $a_{100}$ divides the other $99$ elements.

Mine is case bash. Let us put the elements of $S$ as $a_1,a_2, ... , a_{100}$, such that $a_1 \le a_2\le...\le a_{100}$. $1) a_1\mid a_2,a_3,a_4$ This would give that $a_5$ is either the sum of $3$ numbers divisible by $a_1$ or is directly divisible by $a_1$ which both give $a_1 \mid a_5$ and by induction we prove the claim. $2) a_1$ doesn't divide $a_5$ $a) a_2 \mid a_3,a_4,a_5 \Longrightarrow a_2 \mid a_1 \Longrightarrow a_1=a_2 \Longrightarrow a_1 \mid a_5.\rightarrow \leftarrow$ $b) a_5=a_i+a_j+a_k, 1\le i \le j \le k \le 4$ We see that for a $t \ne i,j,k$ so that $1\le a_t \le 4$ gives $a_t=a_i=a_j=a_k. \rightarrow \leftarrow$

Let $a_{1}$ be the smallest element of S followed by $a_{2}, a_{3}$. Let $a_{k}$ be any arbitrary element such that $k<100$ and $a_{k}=a_{1}+a_{2}+a_{3}$ but then if we consider the set $(a_{1},a_{2},a_{3},a_{100})$ then we get that $a_{1}|a_{2},a_{3}$ and so if we take every 4 element subset containing those 3 numbers ,we are done.If there exist no such k then also we are done. If k=100 then if $a_{100}=a_{1}+a_{2}+a_{3}$ then every number below $a_{100}$ would be divisible by $a_{1}$ also making the sum $a_{1}+a_{2}+a_{3}$ divisible by $a_{1}$. So we are done. I forgot to mention that the elements are in increasing order of their indices.

It is easy to see that there must be 4 elements among the given numbers such that one of these 4 number devides the other 4. Suppose currently we have a set $A$ of $4\le |A|\le 99$ numbers in which one number (call it $x$) devides all the other numbers.Let $x,ax,bx$ be 3 numbers from $A$.For any $t$ not in $A$,take $x,ax,bx$ from $A$ and choose $(x,ax,bx,t)$. If $t$ devides $x$ then $t$ devides all the element of $A$. Otherwise,"sum of three from $(x,ax,bx,t)$ is equal to the other" condition gives that $x|t$. In both cases $A'=A\cup \{t\}$ has property that one element from $A'$ divides all other elements in $A'$.Continue the algorithm untill all the elements are in $A$.

Let $S=\{a_1, a_2, \dots, a_{100}\}$ and $a_1 \le a_2 \le \dots \le a_{100}$. Case 1: There exists $a_k$ such that $a_k = a_1+a_2+a_3$ and $a_k$ appears less than $97$ times. Then if we pick a number $a_j$ such that $a_j \neq a_k$ and $j > 3$ then we know that $a_1$ must divide $a_2, a_3, a_j$. Then if we consider $(a_1, a_2, a_3, a_k)$ then we know that $a_1 \mid a_k$. Then if we consider $(a_1, a_2, a_3, a_i)$ for all $i \ge 4$ we get that $a_1 \mid a_i$. Case 2: There exists $a_k$ such that $a_k = a_1+a_2+a_3$ and $a_k$ appears $97$ times. Consider $(a_1, a_k, a_k, a_k)$. Since none of the numbers is a sum of the other three, that means $a_1 \mid a_k$. Then consider $(a_1, a_2, k, k)$. This means $a_1\mid a_2$. Doing similar thing for $a_3$ we are done. Case 3: There doesn't exist $a_k$ such that $a_k=a_1+a_2+a_3$. Simply taking $(a_1, a_2, a_3, a_i)$ for $i \ge 4$, we are done.