$|A| < 3\sqrt{N}$ when $a < \min A$ divides no nonzero subset sum of $A \subseteq [1,N]$

Let $N \ge 1$, let $a$ be an integer with $1 \le a \le N$, and let $A \subseteq \{1,\ldots,N\}$ be a finite set all of whose elements exceed $a$, i.e. $a < \min A$. Suppose that $a$ divides no nonzero subset sum of $A$: for every nonempty $S \subseteq A$,

Then

This is Corollary 1 of Erdős, Lev, Rauzy, Sándor and Sárközy. It is the arithmetic form of their general group-theoretic Theorem 3, obtained by reading the elements of $A$ modulo $a$: the hypothesis says exactly that the residues form a zero-sum-free sequence in $\mathbb{Z}/a\mathbb{Z}$, and the condition $a < \min A$ together with $A \subseteq \{1,\ldots,N\}$ bounds by $N/a$ the number of elements of $A$ lying in any one residue class modulo $a$.

Applied to $A \setminus \{\min A\}$ with $a = \min A$, it yields the bound $F(N) < 3\sqrt{N}+1$ for non-dividing subsets of $\{1,\ldots,N\}$, and the authors note that it is in fact somewhat stronger than what that application needs. Dropping the hypothesis $a < \min A$ weakens the conclusion to $|A| < 3\sqrt{2N}$ (their Corollary 2).

Formalization Note Membership $a \in \{1,\ldots,N\}$ is kept as an explicit hypothesis, exactly as in the source; it also forces $N \ge 1$, which is what makes the conclusion correct in the degenerate case $A = \emptyset$. The hypothesis $A \subseteq \{1,\ldots,N\}$ is retained although the lower bound $1 \le x$ is already implied by $a < \min A$.