4.C.
$$\begin{array}{l} \text { Given } \frac{1}{a_{n}}=1+\frac{1}{3}+\frac{1}{6}+\cdots+\frac{2}{n(n+1)} \\ =2\left[\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{n(n+1)}\right] \\ =2\left(1-\frac{1}{n+1}\right)=\frac{2 n}{n+1}, \end{array}$$

we know that $a_{n}=\frac{n+1}{2 n}=\frac{1}{2}\left(1+\frac{1}{n}\right)$. Therefore,
$$\begin{array}{l} S_{2000}=\sum_{k=1}^{2000} a_{k}=\frac{1}{2} \sum_{k=1}^{2000}\left(1+\frac{1}{k}\right) \\ =\frac{1}{2} \times 2006+\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2006}\right) . \end{array}$$
$$\begin{array}{l} \text { Also, } 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2006} \\ >1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+ \\ \cdots+(\underbrace{\frac{1}{1024}+\frac{1}{1024}+\cdots+\frac{1}{1024}}_{\text {12 terms }})+ \\ (\underbrace{\frac{1}{2048}+\frac{1}{2048}+\cdots+\frac{1}{2048}}_{\text {22 terms }}) \\ >1+\frac{10}{2}=6, \end{array}$$

then $S_{2000}>1003+3=1006$.
And $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2006}$
$$\begin{array}{l} <1+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right)+ \\ \cdots+(\underbrace{\frac{1}{512}+\frac{1}{512}+\cdots+\frac{1}{512}}_{512 \text{ terms }})+ \\ (\underbrace{\frac{1}{1024}+\frac{1}{1024}+\cdots+\frac{1}{1024}}_{1024 \text{ terms }}) \\ <10+1=11, \end{array}$$

then $S_{2000}<1003+5.5=1008.5$.
Thus, $1006<S_{2006}<1008.5$, which is closest to 1006.