There are $2^{1994}$ such subsets. Let $S = \{1, 2, \dots, 2005\}$ and $R = \{1, 2, 4, 8, \dots, 1024\}$. There are $2^{1994}$ ways to choose a (possibly empty) subset of $S \setminus R$ to include in $B$; whatever sum this has modulo $2048$ will determine uniquely the remainder which the sum of the elements of $B \cap R$ must leave upon division by $2048$. Yet for each such remainder, there is a unique choice of elements in $R$ which sum to it (i.e. those which appear in the binary representation of that number). It follows that the number of possibilities for $B$ is the number of ways to choose a subset of $S \setminus R$, or $2^{1994}$.

I have solution by generating functions:(standard combinatorical way) Let $ \epsilon_k =\cos{\frac{2k\pi}{2048}}+i\sin{\frac{2k\pi}{2048}} $ and $x_m$ be number of subsets with residue is $m$ mod 2048. So \[ x_0+...+x_{2047}\epsilon^{2047}=\prod (1+\epsilon^i). \] We have that $ \prod(1+\epsilon_{j}^{i})=0 $ for all $j=0,1,2,...,2047$. So $x_0=x_1=...=x_{2047}=2^{1994}$.