Three. (25 points) Let $a_{n}$ denote the integer closest to $\sqrt{n}$ (where $n$ is a positive integer). Find the value of $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{200 s}}$.

 untranslated part: 

s should be 00, making it 20000. The problem should read $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{20000}}$. 

However, I will translate the given text as requested:

Three. (25 points) Let $a_{n}$ denote the integer closest to $\sqrt{n}$ (where $n$ is a positive integer). Find the value of $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{200 s}}$.
Three, let $a_{n}=k$, then $|\sqrt{n}-k|<\frac{1}{2}$, which means
$$
\begin{array}{l}
k-\frac{1}{2}<\sqrt{n}<k+\frac{1}{2} \\
\Leftrightarrow k^{2}-k+\frac{1}{4}<n<k^{2}+k+\frac{1}{4} \\
\Leftrightarrow k^{2}-k+1 \leqslant n \leqslant k^{2}+k .
\end{array}
$$

This indicates that when $n$ takes the values $k^{2}-k+1, k^{2}-k+2, \cdots, k^{2}+k$, the integer $a_{n}$ closest to $\sqrt{n}$ is $k$.
Since $44^{2}+44=1980$, we have
$a_{1900}=44$.
Also, $1981=45^{2}-45+1$
$<2005<45^{2}+45=2070$,
thus $a_{1981}=a_{1982}=\cdots=a_{2005}=45$.
Therefore, among $a_{1}, a_{2}, \cdots, a_{2000}$, there are 2 ones, 4 twos, ..., 88 forty-fours, and 25 forty-fives. Hence,
$$
\begin{array}{l}
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2000}} \\
=2 \times \frac{1}{1}+4 \times \frac{1}{2}+\cdots+88 \times \frac{1}{44}+25 \times \frac{1}{45} \\
=44 \times 2+\frac{5}{9}=88 \frac{5}{9} .
\end{array}
$$