jmc

If $\lfloor {\log }_{2}n\rfloor =k$ for some integer $k$, then $k\le {\log }_{2}n<k+1$. Converting to exponential form, this becomes $2^k\le n<2^{k+1}$. Therefore, there are $(2^{k+1}-1)-2^k+1=2^k$ values of $n$ such that $\lfloor {\log }_{2}n\rfloor =k$.

It remains to determine the possible values of $k$, given that $k$ is positive and even. Note that $k$ ranges from $\lfloor {\log }_{2}1\rfloor =0$ to $\lfloor {\log }_{2}999\rfloor =9$. (We have $\lfloor {\log }_{2}999\rfloor =9$ because $2^9\le 999<2^{10}.$) Therefore, if $k$ is a positive even integer, then the possible values of $k$ are $k=2,4,6,8$. For each $k$, there are $2^k$ possible values for $n$, so the answer is $$2^2+2^4+2^6+2^8=\boxed{340}.$$