We claim that such an $ r $ exists if and only if \[\frac{3n^2}{1000} + \frac{3n}{1000^2} + \frac1{1000^3} > 1.\]First, suppose that $ (n+r)^3 $ is an integer, for some $ r \in \left(0, \tfrac{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 $ 3rn^2 + 3nr^2 + r^3 \ge 1 $. Since $ r < \tfrac{1}{1000} $ and $ n>0 $, we get $ \tfrac{3n^2}{1000} + \tfrac{3n}{1000^2} + \tfrac{1}{10^3} > 3rn^2 + 3nr^2 + r^3 \ge 1, $ as desired. Conversely, suppose that $ \tfrac{3n^2}{1000} + \tfrac{3n}{1000^2} + \tfrac{1}{10^3} > 1 $. Define $ f(x) = 3xn^2 + 3nx^2 + x^3 $, so that we have $ f\left(\tfrac{1}{1000}\right) > 1 $. Since $ f(0) = 0 < 1 $ and $ f $ is continuous, there must exist $ r \in \left(0, \tfrac1{1000}\right) $ such that $ f(r) = 1 $. Then for this value of $ r $, we have \[\begin{aligned} (n+r)^3 &= n^3 + 3rn^2 + 3nr^2 + r^3 \\&= n^3 + f(r)\\& = n^3 + 1, \end{aligned}\]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 $ \tfrac{3n^2}{1000} \approx 1 $, or $ n \approx \sqrt{\tfrac{1000}{3}} \approx 18 $. We find that $ n = 18 $ does not satisfy the inequality, but $ n = \boxed{19} $ does.