SOLVED
Let $f(n)$ be maximal such that, for every $m\geq 1$, there exists some $S\subseteq \{1,\ldots,n\}$ with $\lvert S\rvert=f(n)$ such that $m\neq \sum_{a\in A}a$ for all $A\subseteq S$.
Is it true that
\[f(n) = \left(\frac{1}{2}+o(1)\right)\frac{n}{\log n}?\]
A conjecture of Erdős and Graham, who proved the lower bound
\[f(n)\geq \left(\frac{1}{2}+o(1)\right)\frac{n}{\log n}.\]
Their proof is to note that we can assume that $m< \binom{n+1}{2}$ and then, for any $m$, take $S=\{ kp : 1\leq k<\frac{n}{p}\}$ where $p$ is the least prime that does not divide $m$ (so $p<(2+o(1))\log n$).
The complementary bound
\[f(n) \leq \left(\frac{1}{2}+o(1)\right)\frac{n}{\log n}\]
was proved by Alon and Freiman [AlFr88], who chose $m$ as the least common multiple of $\{1,\ldots,s\}$ where $s$ is maximal such that $m\leq \frac{n^2}{20(\log n)^2}$.
