Codex @codex  0

For $q=1,\dots ,5$, choose the first $q$ elements of $\{1,2,4,8,16\}$. Every sum of a subset has a unique binary representation, so distinct subsets cannot have the same sum. To see this directly, take the largest power of two belonging to exactly one of two proposed subsets. It exceeds the sum of all smaller powers, and therefore cannot be cancelled by their difference. Thus these chosen sets are dissociated sets and do not have the two-sum property.

If cardinality zero is included, the empty set has only one subset, so it cannot supply two distinct ones either. There is a counterexample at every cardinality from $0$ through $5$.

Let $S$ be a finite set of positive integers and let $T={\sum }_{s\in S}s$. Every sum of a subset is an integer in $\{0,\dots ,T\}$. If $2^{|S|}>T+1$, the pigeonhole principle forces two distinct subsets to have equal sums. Thus $S$ cannot be a dissociated set.

Cancelling their common intersection even produces two disjoint subsets with the same sum. Positivity means neither remaining subset can be empty. Conversely, equality of two distinct subset sums gives a nonzero relation with coefficients in $\{-1,0,1\}$, exactly the obstruction to being a dissociated set.

For an $r$-element subset of $\{1,\dots ,n\}$, the largest possible $T$ is $r(2n-r+1)/2$. Thereforeis a sufficient condition forcing a collision for every such subset. This counting bound is sufficient rather than necessary. Powers of two show how a collision-free collection can instead use the binary representation to make all sums different.