6. Let $[x]$ denote the greatest integer not exceeding $x$, then $\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\left[\log _{2} 3\right]+\cdots+$ $\left[\log _{2} 2009\right]=$ $\qquad$
6. 17944 Detailed Explanation: When $2^{\prime} \leqslant k<2^{t+1}$, $\left[\log _{2} k\right]=t, t=0,1,2, \cdots$, 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 $2^{10}=1024,2^{11}=2048$, so $\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\left[\log _{2} 3\right]+\cdots+\left[\log _{2} 2009\right]=\sum_{k=1}^{2009} f(k)=f(1)+$ $\sum_{k=2}^{3} f(k)+\sum_{k=4}^{7} f(k)+\sum_{k=8}^{15} f(k)+\sum_{k=16}^{31} f(k)+\sum_{k=32}^{63} f(k)+\sum_{k=64}^{127} f(k)+\sum_{k=128}^{255} f(k)+\sum_{k=256}^{511} f(k)+\sum_{k=512}^{1023} f(k)+$ $\sum_{k=1024}^{2009} f(k)=1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+4 \times 2^{4}+5 \times 2^{5}+6 \times 2^{6}+7 \times 2^{7}+8 \times 2^{8}+9 \times 2^{9}+10 \times\left(2^{10}\right.$ $-39)$. Let $S=1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+4 \times 2^{4}+5 \times 2^{5}+6 \times 2^{6}+7 \times 2^{7}+8 \times 2^{8}+9 \times 2^{9}+10 \times$ $2^{10}$, then $2 S=1 \times 2^{2}+2 \times 2^{3}+3 \times 2^{4}+4 \times 2^{5}+5 \times 2^{6}+6 \times 2^{7}+7 \times 2^{8}+8 \times$ $2^{9}+9 \times 2^{10}+10 \times 2^{11}$, so $S=10 \times 2^{11}-\left[2^{1}+2^{2}+2^{3}+2^{4}+2^{5}+2^{6}+2^{7}+\right.$ $\left.2^{8}+2^{9}+2^{10}\right]=10 \times 2^{11}-2^{11}+2=18334$, so: $\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\left[\log _{2} 3\right]+$ $\cdots+\left[\log _{2} 2009\right]=18334-39 \times 10=17944$.