The answer is $44$.
We shall prove by induction that $\sqrt{n} < a_n < b_n := \frac{n}{\sqrt{n-1}}$ for any integer $n \ge 3$.
For the base case, we find $a_2 = 2$ and $a_3 = 2 \in [\sqrt{3}, \frac{3}{\sqrt{2}}]$.
For the inductive step, assume $\sqrt{k} < a_k < b_k$ for some integer $k \ge 3$. Note that $b_k < k$. Therefore, $\frac{a_k}{k} < \frac{b_k}{k} < 1$. Observe that the function $f(x) = x + \frac{1}{x}$ is

decreasing on $(0,1)$ since $f'(x) = 1 - \frac{1}{x^2} < 0$. Therefore,
$$a_{k+1} = f\left(\frac{a_k}{k}\right) < f\left(\frac{\sqrt{k}}{k}\right) = \frac{k+1}{\sqrt{k}} = b_{k+1}.$$
For the lower bound, we have
$$a_{k+1} = f\left(\frac{a_k}{k}\right) > f\left(\frac{b_k}{k}\right) = \frac{k}{\sqrt{k-1}} > \sqrt{k+1}.$$
So we have proven $\sqrt{n} < a_n < b_n$ for all $n \ge 3$ by induction. This implies
$$a_n < \frac{n}{\sqrt{n-1}} = \sqrt{n+1 + \frac{1}{n-1}} < \sqrt{n+2}.$$
Thus, we have
$$44 < \sqrt{2000} < a_{2000} < \sqrt{2002} < 45.$$
This shows the greatest integer less than or equal to $a_{2000}$ is $44$.