$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$

Since $$\frac1{\sqrt{n+1}}<2(\sqrt{n+1}-\sqrt{n})=\frac2{\sqrt{n+1}+\sqrt{n}}<\frac1{\sqrt{n}},$$ we have $$2(\sqrt{n+1}-\sqrt{2})<\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{1}),$$ i.e. $$2\sqrt{n+1}-2\sqrt{2}+1<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}<2\sqrt{n}-1.$$ For $n=1024$, the bounds are $2\sqrt{1025}-2\sqrt{2}+1\approx62.2$ and $63$, so the result would be $62$.

Use

$$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\in\left(\frac{1}{2\sqrt{n+1}}, \frac{1}{2\sqrt{n}}\right)$$

or, to put it differently:

$$2(\sqrt{n+1}-\sqrt{n})\lt\frac{1}{\sqrt{n}}\lt2(\sqrt{n}-\sqrt{n-1})$$

so you can squeeze your sum between two telescopic sums.

By the EML formula $$ H_n^{(1/2)} = 2\sqrt{n}+\zeta\left(\tfrac{1}{2}\right)+\frac{1}{2\sqrt{n}}+O\left(\frac{1}{n^{3/2}}\right) $$ hence $H_{1024}^{(1/2)}\approx \color{red}{62}+\frac{1}{2}$. There are many questions on MSE asking for approximations of $H_n^{(1/2)}$: the Hermite-Hadamard inequality/the trapezoid method for the estimation of $\int_{1}^{1024}\frac{dx}{\sqrt{x}}$ provide straightforward answers for the task at hand.