China Girls' Mathematical Olympiad

Let $S=\{s_1,s_2,\dots ,s_m\}$. As $S^{\prime }=\{0,s_2-s_1,\dots ,s_m-s_1\}$ satisfies also the hypothesis, we may assume without loss of generality that $0\in S$. For any $x,y\in S$, $50(x+y)\equiv z(\mathrm{mod}99)\in S$. 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 $50^k\equiv 1(\mathrm{mod}99)$. 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 $3$, $9$, $11$, $33$, $99$. $\square$