The *Davenport constant* $D(G)$ of a finite abelian group $G$ is the minimum integer $n$ such that whenever $a_1, \ldots, a_n \in G$ (not necessarily distinct), there is a non-empty $I \subseteq [n]$ such that $\sum_{i \in I} a_i = 0$.

I wonder what happens when we cannot just add the entries up:

Given a finite ground-set, say $[n]$, and a *union-closed* family $\mathcal{A}$ of (non-empty) subsets of $[n]$, let's say $\mathcal{A}$ has the *zero-sum property with respect to $G$* if for every map $\nu : [n] \to G$, there is a $A \in \mathcal{A}$ with $\sum_{x \in A} \nu(x) = 0$.

If $\mathcal{A}$ contains $D(G)$ disjoint sets $A_1, \ldots, A_{D(G)}$, then under any $\nu : [n] \to G$ there will be some $I \subseteq [D(G)]$ such that $\sum_{x \in \bigcup_{i \in I} A_i} v(x) = 0$. Thus $\mathcal{A}$ has the zero-sum property.

If $\mathcal{A}$ can be *covered* by some $X \subseteq [n]$ of size $|X| < D(G)$, meaning that $X$ meets every element of $\mathcal{A}$, then $\mathcal{A}$ does not have the zero-sum property: By definition, there exists a map $\nu : X \to G$ without zero-sum subsequence and we can extend this map by setting $\nu(y) = 0$ for every $y \notin X$.

The question I ultimately want to get at is the following: Does the converse of the last statement also hold, namely: If $\mathcal{A}$ does not have the zero-sum property, does it necessarily admit a cover of size less than $D(G)$? Or, possibly weaker, is there *some* constant $q = q(G)$ such that if $\mathcal{A}$ does not have the zero-sum property, then it admits a cover of size at most $q$?

Using the inclusion-exclusion principle, this is easy to show for $G = \mathbb{Z}_2$, but it is not clear to me how to extend this to other groups.

[The question is related to, but different from this one that I asked on this site a while ago.]

**Edit:** Trimmed the question to make it more accessible.