In a recent Stack Overflow answer of mine, the hardness reduction that I outlined needs a set $U$ of $n$ positive integers such that, given two subsets $S,T\subseteq\binom U3$, if $\sum_{s\in S}s=\sum_{t\in T}t$, then $S=T$.
The obvious solution is $U:=\{2^k\mid k\in[n]\}$, where $[n]:=\{0,1,\ldots,n-1\}$, but then my reduction would only imply weak hardness due to the exponential size of the set members.
In my answer, I used the probabilistic method. Let $U:=\{u_k+1\mid k\in[n]\}$, where $(u_k)_{k\in[n]}\leftarrow[n^6]^n$ is chosen uniformly at random. When $n>3$, it suffices to verify that, for all $J,K\in\binom{[n]}3$, if $\sum_{j\in J}u_j=\sum_{k\in K}u_k$, then $J=K$, since the pairwise distinctness of the $u_k$ follows. There are $\binom{\binom n3}2<\frac{n^6}{12}$ pairs of $J,K$ to worry about, and each contributes a failure with probability less than $n^{-6}$, so the success probability is greater than $\frac{11}{12}$.
This suffices for a Las Vegas reduction, but NP-hardness reductions must be deterministic. Is there an effective polynomial-time construction that uses integers bounded by a polynomial in $n$? I'd also be interested in ways to reduce the polynomial exponent from $6$ (the obvious lower bound is $3$).