For each $i = 1, 2, \dots, 2023$, we have
$$\frac{1}{\sqrt{2023^3 + i}} < \frac{1}{\sqrt{2023^3}} = \frac{1}{2023\sqrt{2023}},$$
thus
$$\frac{1}{\sqrt{2023^3+1}} + \frac{1}{\sqrt{2023^3+2}} + \dots + \frac{1}{\sqrt{2023^3+2023}} < 2023 \cdot \frac{1}{2023\sqrt{2023}} = \frac{1}{\sqrt{2023}}.$$

Similarly, by estimating all fractions from below, we see that
$$\begin{aligned} & \frac{1}{\sqrt{2023^3+1}} + \frac{1}{\sqrt{2023^3+2}} + \dots + \frac{1}{\sqrt{2023^3+2023}} \\ & > 2023 \cdot \frac{1}{\sqrt{2023^3+2023^2}} = \frac{2023}{2023\sqrt{2024}} = \frac{1}{\sqrt{2024}} \end{aligned}$$
Therefore, $a = 2023^3$ is suitable.