［Solution］If we set $2^{k} \leqslant m<2^{k+1}$, then $\left[\log _{2} m\right]=k$.
Also, in the interval $\left[2^{k}, 2^{k+1}\right)$, there are $2^{k}$ integers.
$$
\begin{aligned}
\therefore \quad & \sum_{N=1}^{1024}\left[\log _{2} N\right]=\sum_{N=1}^{2^{10}}\left[\log _{2} N\right] \\
= & 1 \cdot 2^{1}+2 \cdot 2^{2}+3 \cdot 2^{3}+4 \cdot 2^{4}+\cdots+9 \cdot 2^{9}+10
\end{aligned}
$$

Let the sum on the right side of the above equation be $s$, consider
$$
\begin{aligned}
2 s-s & =10+9 \cdot 2^{10}-2^{9}-2^{8}-\cdots-2^{1} \\
& =10+9 \cdot 2^{10}-\left(2^{10}-2\right),
\end{aligned}
$$
$$
s=8204 \text {. }
$$

Therefore, the answer is $(B)$.
33.42 If $[x]$ is the greatest integer less than or equal to $x$, then $\sum_{N=1}^{1024}\left[\log _{2} N\right]$ equals
(A) 8192.
(B) 8204.
(C) 9218.
(D) $\left[\log _{2}(1024!)\right]$.
(E) None of the above.
(37th American High School Mathematics Examination, 1986)