Solution:
We wish to find $n$ such that there is some natural number $k$ for which $2k-1 \leq \log_{2} n < 2k$. Since $n \leq 2009$ we must have $k \leq 5$. This is equivalent to finding the number of positive integers $n \leq 2009$ satisfying $2^{2k-1} \leq n < 2^{2k}$ for some $k \leq 5$, so the number of such integers is $2 + 2^{3} + 2^{5} + 2^{7} + 2^{9} = 682$.