Let $m$ be the smallest [integer](https://artofproblemsolving.com/wiki/index.php/Integer) whose [cube root](https://artofproblemsolving.com/wiki/index.php?title=Cube_root&action=edit&redlink=1) is of the form $n+r$, where $n$ is a [positive integer](https://artofproblemsolving.com/wiki/index.php/Positive_integer) and $r$ is a [positive](https://artofproblemsolving.com/wiki/index.php/Positive) [real number](https://artofproblemsolving.com/wiki/index.php/Real_number) less than $1/1000$. Find $n$.
In order to keep $m$ as small as possible, we need to make $n$ as small as possible.
$m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3$.  Since $r  1000$.  This means that the smallest possible $n$ should be less than 1000. In particular, $3nr + r^2$ should be less than 1, so $3n^2 > 999$ and $n > \sqrt{333}$.  $18^2 = 324  3\cdot 19^2 > 1000$, 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  $\boxed{019}$.