Is this year 2025? I don't think so!

We can estimate the value of $\displaystyle S = \sum_{k=1}^{n^2} \sqrt{\frac 1k}$ with $\displaystyle I = \int_a^b \sqrt{\frac 1x} dx$ with appropriate values of the limits $a$ and $b$ . Since the $\dfrac 1{\sqrt x}$ is convex, we note that the area where the curve is above the column $\alpha$ is larger than the area where the column is above the curve $\beta$ . Then we have:

$\begin{aligned} \frac 12 + \int_1^{n^2} \frac 1{\sqrt x} dx + \frac 1{\sqrt{n + \frac 12}} & < \sum_{k=1}^{n^2} \sqrt{\frac 1k} < \int_\frac 12^{n^2 + \frac 12} \frac 1{\sqrt x} dx \\ \frac 12 + 2\sqrt x \ \bigg|_1^{n^2} + \frac 1{\sqrt{n + \frac 12}} & < \sum_{k=1}^{n^2} \sqrt{\frac 1k} < \int_\frac 12^1 \frac 1{\sqrt x} dx + \int_1^{n^2} \frac 1{\sqrt x} dx + \color{#3D99F6} \int_{n^2}^{n^2 + \frac 12} \frac 1{\sqrt x} dx & \small \color{#3D99F6} \text{Note that }\int_{n^2}^{n^2 + \frac 12} \frac 1{\sqrt x} dx < \frac 12 \cdot \frac 1{\sqrt{n^2}} = \frac 1{2n} \\ \frac 12 + 2(n-2) + \frac 1{\sqrt{n + \frac 12}} & < \sum_{k=1}^{n^2} \sqrt{\frac 1k} < 2-\sqrt 2 + 2(n-1) + \color{#3D99F6} \frac 1{2n} & \small \color{#3D99F6} \text{one half of the last column.} \\ \left \lfloor \frac 12 + 2(n-2) + \frac 1{\sqrt{n + \frac 12}} \right \rfloor & < \left \lfloor \sum_{k=1}^{n^2} \sqrt{\frac 1k} \right \rfloor < \left \lfloor 2-\sqrt 2 + 2(n-1) + \frac 1{2n} \right \rfloor & \small \color{#3D99F6} \text{Taking the integral part throughout.} \\ \implies 2(n-1) & = \left \lfloor \sum_{k=1}^{n^2} \sqrt{\frac 1k} \right \rfloor = 2(n-1) & \small \color{#3D99F6} \text{for } n \ge 2 \end{aligned}$

Therefore $\displaystyle \left \lfloor \sum_{k=1}^{2025} \sqrt{\frac 1k} \right \rfloor = 2(\sqrt{2025}-1) = \boxed{88}$ .