6. Let set $S$ be an $m(m \geqslant 3)$-element subset of $\{0,1, \cdots, 98\}$, satisfying that for any $x, y \in S$, there exists $z \in S$ such that $x+y \equiv 2 z(\bmod 99)$. Find all possible values of $m$.
(Wang Xinmao provided the problem)
6. Let $S=\left\{s_{1}, s_{2}, \cdots, s_{m}\right\}$.

Since $S^{\prime}=\left\{0, s_{2}-s_{1}, \cdots, s_{m}-s_{1}\right\}$ also satisfies the condition, we may assume without loss of generality that $0 \in S$.
By the condition, for any $x, y \in S$,
$50(x+y) \equiv z(\bmod 99) \in S$.
In particular, for any $x \in S$, we have $50 x \in S$.
By $(50,99)=1$ and Euler's theorem, there exists a positive integer $k$ such that
$50^{k} \equiv 1(\bmod 99)$.
Thus, $x+y \equiv 50^{k}(x+y)(\bmod 99) \in S$.
Let $d=\left(99, s_{1}, s_{2}, \cdots, s_{m}\right)$. Then by the above conclusion and Bézout's theorem, we know that $d \in S$.
Therefore, $S=\{0, d, 2 d, \cdots\}$.
For any positive divisor $d<99$ of 99, the above set $S$ satisfies the condition.
Hence, all possible values of $m$ are $3,9,11,33,99$.