In short: In how many ways can all $2^n$ subset sums of $n$ real numbers $a_1,\ldots, a_n$ be ordered? I am not concerned about the case in which different subsets sum to the same number; you may assume all $2^n$ subset sums are distinct.
Example: Suppose $n=2$. There are four subset sums: $0$, $a_1$, $a_2$, and $a_1+a_2$. There are eight possible orderings: \begin{eqnarray} 0 &<& a_1 < a_2 < a_1 + a_2\\ 0 &<& a_2 < a_1 < a_1 + a_2\\ a_1 &<& 0 < a_1 + a_2 < a_2 \\ a_2 &<& 0 < a_1 + a_2 < a_1 \\ a_1 &<& a_1 + a_2 < 0 < a_2 \\ a_2 &<& a_1 + a_2 < 0 < a_1 \\ a_1 + a_2 &<& a_2 < a_1 < 0 \\ a_1 + a_2 &<& a_1 < a_2 < 0. \end{eqnarray} All other orderings of these four expressions, such as $$ a_1 + a_2 < 0 < a_1 < a_2 $$ are not possible.
Motivation: This problem is directly related to the well-known NP-complete "subset sum" problem in computer science. I am wondering whether the pure combinatorics of that problem render it especially difficult.
My guess is that the problem I pose here has been asked before: I think it's a pretty straightforward way to understand something about the difficulty of the subset sum problem.