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 \[2^2 + 2^4 + 2^6 + 2^8 = \boxed{340}.\]