Start Counting

We consider the elements in $S$ modulo $100$ . There are $100$ congruence classes mdoulo $100$ , namely

$0,1,2,\ldots,99$

If $S$ contains at least $101$ elements, by the Pigeonhole principle , there must be two elements $x$ and $y$ that belong to the same congruence class, in other words

$\begin{aligned} &x\equiv y \mod 100\\ \iff &x-y\equiv 0\mod 100\\ \iff & 100\mid x-y \end{aligned}$

$100$ elements are not enogh, since we could have that every element belongs to a different class. Hence the minimum number of elements that satisfy the constraint is $\boxed{101}$ .

More generally if the problem asked the same question with $n\mid x-y$ the answer would be $n+1$