China Girls' Mathematical Olympiad

Let $S = \{s_1, s_2, \dots, s_m\}$. As $S' = \{0, s_2 - s_1, \dots, s_m - s_1\}$ satisfies also the hypothesis, we may assume without loss of generality that $0 \in S$. For any $x, y \in S$, $50(x + y) \equiv z \pmod{99} \in S$. By taking $y = 0$, we have that for any $x \in S$, $50x \in S$. As $50$ and $99$ are coprime, there exists a positive integer $k$ such that $50^k \equiv 1 \pmod{99}$. Hence,
$$x + y \equiv 50^k (x + y) (\text{mod } 99) \in S.$$
Let $d = \gcd(99, s_1, s_2, \dots, s_m)$. The above argument then implies that $d \in S$, therefore $S = \{0, d, 2d, \dots\}$. For any positive factor $d < 99$ of $99$, this $S$ satisfies all the requirements. Hence, all the possible values of $m$ are $3$, $9$, $11$, $33$, $99$. $\square$