8. Arrange all positive integers that leave a remainder of 2 and 3 when divided by 4 in ascending order. Let $S_{n}$ denote the sum of the first $n$ terms of this sequence. Then $\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2012}}\right]$ $=$ $\qquad$ ([ $x]$ denotes the greatest integer not exceeding the real number $x$).
8. 2025078.

Given that this sequence is
$$
2,3,6,7, \cdots, 4 n-2,4 n-1, \cdots \text {. }
$$

From $4 n-2+4 n-1=8 n-3$, we know
$$
\begin{array}{l}
S_{2 n}=5+13+\cdots+(8 n-3) \\
=4 n(n+1)-3 n=4 n^{2}+n . \\
\text { Also, } 4 n^{2}<S_{2 n}<(2 n+1)^{2} \text {, then } \\
2 n<\sqrt{S_{2 n}}<2 n+1 .
\end{array}
$$

Therefore, $\left[S_{2 n}\right]=2 n$.
Also, $S_{2 n-1}=S_{2 n}-(4 n-1)=4 n^{2}-3 n+1$, then $(2 n-1)^{2}<S_{2 n-1}<4 n^{2}$.
Therefore, $\left[S_{2 n-1}\right]=2 n-1$.
$$
\begin{array}{l}
\text { Hence }\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2012}}\right] \\
=1+2+\cdots+2012=2025078 .
\end{array}
$$