Solution:

Let $a < b$ be the two smallest numbers of $S$ and let $d$ be the largest number of $S$. Consider any two other numbers $x < y$ of $S$. For the quadruples $(a, b, x, d)$ and $(a, b, y, d)$ we cannot get both of $d = a + b + x$ and $d = a + b + y$, since $a + b + x < a + b + y$. From here, we get $a \mid b$ and $a \mid d$.

Consider any number $s$ of $S$ different from $a, b, d$. From the condition of the problem, we get $d = a + b + s$ or $a$ divides $b, s$ and $d$. But since we already know that $a$ divides $b$ and $d$ anyway, we also get that $a \mid s$, as in the first case we have $s = d - a - b$. This means that $a$ divides all other numbers of $S$.

Solution:

Order the elements of $S$ as $x_{1} < x_{2} < \cdots < x_{100}$.

For $2 \leqslant k \leqslant 97$, looking at the quadruples $(x_{1}, x_{k}, x_{k+1}, x_{k+2})$ and $(x_{1}, x_{k}, x_{k+1}, x_{k+3})$, we get that $x_{1} \mid x_{k}$ as alternatively, we would have $x_{k+2} = x_{1} + x_{k} + x_{k+1} = x_{k+3}$, a contradiction.

For $5 \leqslant k \leqslant 100$, looking at the quadruples $(x_{1}, x_{k-2}, x_{k-1}, x_{k})$ and $(x_{1}, x_{k-3}, x_{k-1}, x_{k})$ we get that $x_{1} \mid x_{k}$ as alternatively, we would have $x_{k} = x_{1} + x_{k-2} + x_{k-1} = x_{1} + x_{k-3} + x_{k-1}$, a contradiction.

So $x_{1}$ divides all other elements of $S$.

Solution:

The condition that one element is the sum of the other three cannot be satisfied by all quadruples. So we have four elements such that one divides the other three. Suppose inductively that we have a subset $S'$ of $S$ with $|S'| = k \geqslant 4$ such that there is $x \in S'$ with $x \mid y$ for every $y \in S'$. Pick $s \in S \setminus S'$ and $y, z \in S'$ different from $x$. Considering $(s, x, y, z)$ either $s \mid x$, or $x \mid s$ or one of the four is a sum of the other three. In the last case we have $s = \pm x \pm y \pm z$ and so $x \mid s$. In any case either $x$ or $s$ divides all elements of $S' \cup \{s\}$.