Can $\sum_{n=1}^{k} \frac1{\sqrt{n}}$ be an integer for $k>1$? In one of new YouTube videos, I've seen the problem: to show that
$$16 < \sum_{n=1}^{80} \frac1{\sqrt{n}} < 17$$
I've solved the problem easily, using
$$\int_2^{81}\frac1{\sqrt{x}}\, dx < \sum_{n=2}^{80} \frac1{\sqrt{n}} < \int_1^{81}\frac1{\sqrt{x}}\, dx$$
But this problem aimed me on another problem:

Can the sum $\sum_{n=1}^{k}  \frac1{\sqrt{n}}$ be an integer number at some positive integer $k>1$?

Is this problem well-known open or closed problem? Or maybe there is some approach allowing to solve this problem easily without using hard mathematical skills?
I've seen the same problem without square roots and it was solved easily.

This is not an integer due to the same reason that $H_n$ is not an integer.
Take the largest integer $k$ such that $2^k < n$. If $M = 2^{k - 1} \cdot (2n - 1)!!$, then $\frac{\sqrt{M}}{\sqrt{i}}$ is an algebraic integer for any $i \neq 2^k$ between $1$ and $n$, but $\frac{\sqrt{M}}{\sqrt{2^k}}$ is not algebraic integer. So
$$\sqrt{M} \cdot \sum_{i = 1}^n \frac{1}{\sqrt{i}}$$
is not algebraic integer.
