jmc

In order to keep $m$ as small as possible, we need to make $n$ as small as possible. $m=(n+r{)}^3=n^3+3n^{2}r+3n{r}^2+r^3$. Since $r<\frac{1}{1000}$ and $m-n^3=r(3n^2+3nr+r^2)$ is an integer, we must have that $3n^2+3nr+r^2\ge \frac{1}{r}>1000$. This means that the smallest possible $n$ should be quite a bit smaller than 1000. In particular, $3nr+r^2$ should be less than 1, so $3n^2>999$ and $n>\sqrt{333}$. $18^2=324<333<361=19^2$, so we must have $n\ge 19$. Since we want to minimize $n$, we take $n=19$. Then for any positive value of $r$, $3n^2+3nr+r^2>3\cdot 19^2>1000$, so it is possible for $r$ to be less than $\frac{1}{1000}$. However, we still have to make sure a sufficiently small $r$ exists. In light of the equation $m-n^3=r(3n^2+3nr+r^2)$, we need to choose $m-n^3$ as small as possible to ensure a small enough $r$. The smallest possible value for $m-n^3$ is 1, when $m=19^3+1$. Then for this value of $m$, $r=\frac{1}{3n^2+3nr+r^2}<\frac{1}{1000}$, and we're set. The answer is $\boxed{19}$.