Solution:
Let us consider the set $\{1,2,2^{2}, \ldots, 2^{10}\}$. Since every number from $0$ to $2047$ can be represented in a unique way as a sum of powers of $2$ (elements of our set), we conclude that for every $i$, $0 \leq i \leq 2047$, there is a unique subset of $\{1,2,2^{2}, \ldots, 2^{10}\}$ such that the sum of its elements is equal to $i$ ($0$ corresponds to the empty set).

We now consider a set $A$ with the following property: for every $i$ the number of the subsets of $A$ such that the sums of their elements are congruent to $i$ modulo $2048$ does not depend on $i$. It is easy to see that for every $a$ the set $A \cup\{a\}$ has the same property. Since $\{1,2,2^{2}, \ldots, 2^{10}\} \subset \{1,2,3, \ldots, 2005\}$, we conclude that the number of the subsets $B$ of $\{1,2,3, \ldots, 2005\}$ such that the sums of their elements are congruent to $i$ modulo $2048$ is equal to $\frac{2^{2005}}{2048} = \frac{2^{2005}}{2^{11}} = 2^{1994}$ and this is the required number.