(9) Let $[x]$ denote the greatest integer not exceeding $x$, then $\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\left[\log _{2} 3\right]+\cdots$ $+\left[\log _{2} 500\right]=$ $\qquad$ .
(9) When $2^{t} \leqslant k<2^{t+1}$, $\left[\log _{2} k\right]=t(t=0,1,2, \cdots)$, and there are $2^{t}$ positive integers in the interval $\left[2^{t}, 2^{t+1}\right)$. Let $f(x)=\left[\log _{2} x\right]$, noting that $2^{9}=512$, so
$$
\begin{aligned}
& {\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\left[\log _{2} 3\right]+\cdots+\left[\log _{2} 500\right] } \\
= & \sum_{k=1}^{500} f(k) \\
= & f(1)+\sum_{k=2}^{3} f(k)+\sum_{k=4}^{7} f(k)+\sum_{k=8}^{15} f(k)+\sum_{k=16}^{31} f(k)+\sum_{k=32}^{63} f(k) \\
& +\sum_{k=64}^{127} f(k)+\sum_{k=128}^{255} f(k)+\sum_{k=256}^{500} f(k) \\
= & 0+1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+4 \times 2^{4}+5 \times 2^{5}+6 \times 2^{6}+7 \times 2^{7} \\
& +8 \times\left(2^{8}-11\right) .
\end{aligned}
$$

Let
$$
\begin{aligned}
S= & 1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+4 \times 2^{4}+5 \times 2^{5} \\
& +6 \times 2^{6}+7 \times 2^{7}+8 \times 2^{8},
\end{aligned}
$$

then
$$
\begin{aligned}
2 S= & 1 \times 2^{2}+2 \times 2^{3}+3 \times 2^{4}+4 \times 2^{5}+5 \times 2^{6} \\
& +6 \times 2^{7}+7 \times 2^{8}+8 \times 2^{9},
\end{aligned}
$$

subtracting the two equations gives
$$
\begin{aligned}
-S & =2^{1}+2^{2}+2^{3}+2^{4}+2^{5}+2^{6}+2^{7}+2^{8}-8 \times 2^{9} \\
& =2^{9}-2-8 \times 2^{9},
\end{aligned}
$$

so
$$
S=7 \times 2^{9}+2=3586 .
$$

Therefore
$$
\begin{aligned}
& {\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\left[\log _{2} 3\right]+\cdots+\left[\log _{2} 500\right] } \\
= & 3586-8 \times 11=3498 .
\end{aligned}
$$