9. (1990 Hungarian Mathematical Olympiad) For any positive integer $q_{0}$, consider the sequence $q_{1}, q_{2}, \cdots, q_{n}$ defined by $q_{i}=\left(q_{i-1}-1\right)^{3}+$ $3(i=1,2, \cdots, n)$. If each $q_{i}(i=1,2, \cdots, n)$ is a prime, find the largest possible value of $n$.
9. Since $m^{3}-m=m(m-1)(m+1) \equiv 0(\bmod 3)$, we have
$$
q_{i}=\left(q_{i-1}-1\right)^{3}+3 \equiv\left(q_{i-1}-1\right)^{3} \equiv q_{i-1}-1(\bmod 3) \text {. }
$$

Therefore, among $q_{1}, q_{2}, q_{3}$, there must be one that is divisible by 3, and this number should be a power of 3.
If $3 \mid(q-1)^{3}+3$, then $3 \mid(q-1)^{3}$, so $3 \mid q-1$.
Hence, $3^{3} \mid(q-1)^{3}$.
And since $3 \mid(q-1)^{3}+3$,
it follows that $\left(q_{i}-1\right)^{3}+3$ is a power of 3 only when $q_{i}=1$, which must occur when $i=0$.
However, when $q_{0}=1$, we get
$$
q_{1}=3, q_{2}=11, q_{3}=1003=17 \cdot 59 \text {. }
$$

Thus, the maximum value of $n$ is 2.