smc

Because $1\le N\le 1024$, we have $0\le \lfloor {\log }_{2}N\rfloor \le 10$. We count how many times $\lfloor {\log }_{2}N\rfloor$ attains a certain value. For all $k$ except for $k=10$, we have that $\lfloor {\log }_{2}N\rfloor =k$ is satisfied by all $2^k\le N<2^{k+1}$, for a total of $2^k$ values of $N$. If $k=10$, $N$ can only have one value ($N=1024$). Thus, the desired sum looks like $$\sum _{N=1}^{1024}\lfloor {\log }_{2}N\rfloor =1(0)+2(1)+4(2)+\cdots +2^k(k)+\cdots +2^9(9)+10$$ Let $S$ be the desired sum without the $10$. $$S=2(1)+4(2)+\cdots +2^9(9)$$ Multiplying by $2$ gives $$2S=4(1)+8(2)+\cdots +2^{10}(9)$$ Subtracting the two equations gives $$S=2^{10}(9)-(2+4+8+\cdots +2^9)$$ Summing the geometric sequence in parentheses and simplifying, we get $$S=2^{10}(9)-2^{10}+2=2^{10}(8)+2=8194$$ Finally, adding back the $10$ gives the desired answer $\boxed{8204}$