8. (26th Russian Mathematical Olympiad) Find the sum $\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{2^{2}}{3}\right]+\left[\frac{2^{3}}{3}\right]+\cdots+\left[\frac{2^{1000}}{3}\right]$.
8. The first term is obviously 0, so we discard it. For the remaining 1000 terms, we first examine
$$
\frac{2}{3}+\frac{2^{2}}{3}+\frac{2^{3}}{3}+\cdots+\frac{2^{1000}}{3} \text {. }
$$

This is the sum of a geometric series, whose value is $\frac{1}{3}\left(2^{1001}-2\right)$. Next, we examine the sum of their integer parts. In the above series, no term is an integer, but the sum of any two consecutive terms is an integer (since $\frac{2^{k}}{3}+\frac{2^{k-1}}{3}=\frac{3 \cdot 2^{k}}{3}=2^{k}$). It is easy to see that if the sum of two non-integers is an integer, then the sum of their integer parts is 1 less than their sum ($\alpha+\beta=[\alpha]+\{a\}+[\beta]+\{\beta\}=[\alpha]+[\beta]+\{\alpha\}+\{\beta\}=[\alpha]+[\beta]+1$).

Therefore, once we replace every two consecutive terms in the geometric series with their integer parts, their sum decreases by 1. Since there are 1000 terms in the series, the sum of their integer parts is 500 less than the original sum. Hence, the required sum is $\frac{1}{3}\left(2^{1001}-2\right)-500$.