12. The square root of a number without $m$ is $\cdots$ numbers of the form $n+r$, where $n$ is a positive integer, and $r$ is a real number less than $\frac{1}{1000}$. If $m$ is the smallest positive integer satisfying the above condition, find the value of $n$ when $m$ is the smallest positive integer.
12. From the problem: $\sqrt[3]{m}=n+r, 0\frac{1000}{3}
\end{array}
$
$\rightarrow n^{2} \approx \frac{1000}{3}$. Since $18^{2}<\frac{1000}{3}<19^{2}$, we can guess $n=18$ or $n=19$. Upon verification, when $n=18$, the inequality does not hold, but when $n=19$, the inequality is satisfied. Therefore, $n=19$ is the smallest positive integer that meets the problem's requirements (at this time, $m=19^{3}+1=6860$ is the smallest positive integer whose cube root has a positive fractional part less than $\frac{1}{1000}$).