Example 9 Find the smallest positive integer $n$, such that the last three digits of $n^{3}$ are 888.
Solution: Since the unit digit of $n^{3}$ is 8, then the unit digit of $n$ is 2. Let $n=10 k+2$. Then
$$
\begin{array}{l}
n^{3}=(10 k+2)^{3}=1000 k^{3}+600 k^{2}+120 k+8 \\
=1000 k^{3}+100 k(6 k+1)+20 k+8 .
\end{array}
$$

According to the problem, we know that the last three digits of $100\left(6 k^{2}+k\right)+20 k+8$ are 888, so the last two digits of $20 k+8$ are 88.
Therefore, $20 k=100 m+80$, which means $k=5 m+4$.
Thus, $100\left(6 k^{2}+k\right)+20 k+8$
$$
\begin{array}{l}
=100(5 m+4)(30 m+25)+20(5 m+4)+8 \\
=15000 m^{2}+24600 m+10088 .
\end{array}
$$

Therefore, the unit digit of $6 m$ is 8. Hence, the smallest value of $m$ is 3. Consequently, the smallest value of $k$ is 19. Therefore, the smallest value of $n$ is 192.