Don't give up, try this

We note that:

$\displaystyle \sum_{k=\color{#D61F06}{2}}^{80} {\frac{1}{\sqrt{k}}} < \int_1^{80} {\frac{1}{\sqrt{x}}dx} < \sum_{k=\color{#3D99F6}{1}}^{80} {\frac{1}{\sqrt{k}}} - \frac{1}{2}\times \frac{1}{\sqrt{1}} - \frac{1}{2}\times \frac{1}{\sqrt{80}} \\ S - \frac{1}{\sqrt{1}} < \int_1^{80} {\frac{1}{\sqrt{x}}dx} < S - \frac {\sqrt{80}+1}{2\sqrt{80}}\\ S - 1 < \left[ 2\sqrt{x} \right]_1^{80} < S - \frac {\sqrt{80}+1}{2\sqrt{80}} \\ S - 1 < 2\left(\sqrt{80} - 1\right) < S - \frac {\sqrt{80}+1}{2\sqrt{80}}\\ S - 1 < 15.88854382 < S - 0.555901699 \\ 16.44444552 < S <16.88854382 \\ \Rightarrow \lfloor S \rfloor = \boxed{16}$

This isn't really a proper solution, but what I did to get close enough to get it was to approximate the value between "nice" points using a few different methods. For example, for every value between $k=9$ and $k=16$, the terms lie between $1/3$ and $1/4$. After trying with left, right and midpoint, and summing with a hand calculator, I found the answer to be 16, and this was correct. I am looking forward to seeing someone more creative than me find a more analytic solution though!