jmc

We claim that such an $r$ exists if and only if $$\frac{3n^2}{1000}+\frac{3n}{1000^2}+\frac{1}{1000^3}>1.$$First, suppose that $(n+r{)}^3$ is an integer, for some $r\in \left(0,\frac{1}{1000}\right).$ Since $(n+r{)}^3>n^3$ and $n^3$ is an integer, we must have $$(n+r{)}^3\ge n^3+1,$$so $3r{n}^2+3n{r}^2+r^3\ge 1.$ Since $r<\frac{1}{1000}$ and $n>0$, we get $\frac{3n^2}{1000}+\frac{3n}{1000^2}+\frac{1}{10^3}>3r{n}^2+3n{r}^2+r^3\ge 1,$ as desired.

Conversely, suppose that $\frac{3n^2}{1000}+\frac{3n}{1000^2}+\frac{1}{10^3}>1.$ Define $f(x)=3x{n}^2+3n{x}^2+x^3$, so that we have $f\left(\frac{1}{1000}\right)>1.$ Since $f(0)=0<1$ and $f$ is continuous, there must exist $r\in \left(0,\frac{1}{1000}\right)$ such that $f(r)=1.$ Then for this value of $r$, we have $$\begin{array}{rl}(n+r{)}^3& =n^3+3r{n}^2+3n{r}^2+r^3\\ & =n^3+f(r)\\ & =n^3+1,\end{array}$$which is an integer, as desired.

Thus, it suffices to find the smallest positive integer $n$ satisfying $$\frac{3n^2}{1000}+\frac{3n}{1000^2}+\frac{1}{1000^3}>1.$$The first term on the left-hand side is much larger than the other two terms, so we look for $n$ satisfying $\frac{3n^2}{1000}\approx 1$, or $n\approx \sqrt{\frac{1000}{3}}\approx 18$. We find that $n=18$ does not satisfy the inequality, but $n=\boxed{19}$ does.