Estonian Mathematical Olympiad

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 $$\frac{1}{\sqrt{2023^3+1}}+\frac{1}{\sqrt{2023^3+2}}+\cdots +\frac{1}{\sqrt{2023^3+2023}}<2023\cdot \frac{1}{2023\sqrt{2023}}=\frac{1}{\sqrt{2023}}.$$

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