8. Remove all positive integers that are divisible by 4 and those that leave a remainder of 1 when divided by 4, and arrange the remaining numbers in ascending order to form a sequence $\left\{a_{n}\right\}$:
$$
2,3,6,7,10,11, \cdots \text {. }
$$

The sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is denoted as $S_{n}$, where $n=$ $1,2, \cdots$. Find
$$
S=\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2000}}\right]
$$

where $[x]$ denotes the greatest integer not exceeding $x$.
8. It is known that $a_{2 n-1}=4 n-2, a_{2 n}=4 n-1(n=1,2$, $\cdots$ ). Therefore,
$$
\begin{aligned}
& S_{2 n}=\left(a_{1}+a_{2}\right)+\left(a_{3}+a_{4}\right)+\cdots+\left(a_{2 n-1}+a_{2 n}\right) \\
= & 5+13+21+\cdots+(8 n-3) \\
= & \frac{(5+8 n-3) n}{2}=(2 n)^{2}+n, \\
& S_{2 n-1}=S_{2 n}-a_{2 n}=4 n^{2}+n-(4 n-1) \\
= & (2 n-1)^{2}+n .
\end{aligned}
$$

Thus, $(2 n)^{2}<S_{2 n}<(2 n+1)^{2}$,
$$
\begin{array}{l}
(2 n-1)^{2}<S_{2 n-1}<(2 n)^{2} . \\
\text { Hence }\left[\sqrt{S_{2 n}}\right]=2 n,\left[\sqrt{S_{2 n-1}}\right]=2 n-1 .
\end{array}
$$

Therefore, $\left[\sqrt{S_{n}}\right]=n$. Thus,
$$
\begin{array}{l}
S=\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2000}}\right] \\
=1+2+\cdots+2006 \\
=\frac{2006 \times 2007}{2}=2013021 .
\end{array}
$$