9. Find the smallest positive integer $n$, such that the last three digits of its cube are 888.
9. (192)

If a positive integer ends with 8, then this number itself must end with 2. That is, it can be written in the form $10k + 2$. Thus,
$$
n^{3} = (10k + 2)^{3} = 1000k^{3} + 600k^{2} + 120k + 8.
$$

The term $120k$ determines the tens digit of $n^{3}$, which is 8. Therefore, $12k$ must end with 8, meaning $k$ is 4 or 9. Thus, $k = 5m + 4$. ($m$ is a non-negative integer). Then,
$$
\begin{aligned}
n^{3} = & {[10(5m + 4) + 2]^{3}} \\
= & 125000m^{3} + 315000m^{2} + 26460m \\
& + 70488.
\end{aligned}
$$

Since the last three digits of the first two terms on the right side are 000, and the last three digits of the last term are 088, the last three digits of $26460m$ must be 800. The smallest integer $m$ that satisfies this condition is 3. Thus, $k = 5 \times 3 + 4 = 19$, and $n = 10 \times 19 + 2 = 192$.