Cute problem, wonder why no one has posted a solution yet. Consider the set of powers of two that appear in the binary expansions of the $a_i$. We partition this set into $n \leq 2^m-1$ parts $B_1, B_2, \dots, B_n$, and let $b_k$ be the sum of elements in the $B_k$. Clearly, this ensures that all sums of distinct $b_k$'s are distinct. Now we have to ensure that, for each $a_i$, the set of powers of two that appear in the binary expansion of $a_i$ can be written as a union of some $B_j$. Indeed, index the $B_j$ by the non-empty subsets of $\{1,2, \dots, m \}$, and for any such subset $S$, $B_S$ is precisely the set of powers of two that appear in binary expansion $a_i$ for $i \in S$ and don't appear in binary expansion of $a_j$ for $j \notin S$ (only the $S$ for which $B_S$ is non-empty are counted). Then the set of powers of two that appear in binary expansion of $a_i$ is precisely the union of $B_S$ for all sets $S$ containing $i$, as required.