What is the smallest positive integer $n$ for which $\sqrt{2019-n}$ is an integer?
In order for $\sqrt{2019-n}$ to be an integer, $2019-n$ must be a perfect square.

Since $n$ is a positive integer, then $2019-n$ is a perfect square less than 2019 .

Since $n$ is to be as small as possible, $2019-n$ must be the largest perfect square less than 2019 . Since $44^{2}=1936$ and $45^{2}=2025$, then 1936 is the largest perfect square less than 2019.

Thus, $2019-n=1936$ and so $n=83$.

ANSWER: 83