Just Inequality, No More

Since $\tfrac{1}{\sqrt{x}}$ is a decreasing function of $x$ , we have $\sum_{k=1}^{80} \frac{1}{\sqrt{k}} \; \ge\; 1 + \int_2^{81} \frac{dx}{\sqrt{x}} \; = \; 1 + 2(\sqrt{81} - \sqrt{2}) = 16.117157288$ while $\sum_{k=1}^{80} \frac{1}{\sqrt{k}} \; \le \; 1 + \int_1^{80} \frac{dx}{\sqrt{x}} \; = \; 1 + 2(\sqrt{80}-1) \; = \; 16.88854382$ and hence $\left\lfloor \sum_{k=1}^{80} \frac{1}{\sqrt{k}} \right\rfloor \; = \; \boxed{16}$