6-102 Let the cube root of $m$ be a number of the form $n+r$, where $n$ is a positive integer, and $r$ is a positive real number less than $\frac{1}{1000}$. When $m$ is the smallest positive integer satisfying the above condition, find the value of $n$.

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[Solution] According to the problem, $\sqrt[3]{m}=n+r$, where $n$ is a positive integer, $0 < r < 1$, and $m$ is a positive integer. Therefore, we have:
$$
\begin{array}{l}
(n+r)^{3}=m \\
n^{3} < m < (n+1)^{3} \\
n^{3} < \frac{1000}{3} < (n+1)^{3}
\end{array}
$$

Thus, $n^{2} \approx \frac{1000}{3}$,
which means $n^{2}=18^{2}$ or $19^{2}$.
Upon verification, when $n^{2}=18^{2}$, equation (1) does not hold; whereas when $n^{2}=19^{2}$, equation (1) holds.
Therefore, $n=19$ is the smallest positive integer that satisfies the problem.