Looking for a counterexample to a strengthening of the union-closed sets conjecture

Question

[Now crossposted at math.stackexchange]

Let $\mathcal{F} = \{\{x_1, x_2\} : 1 \le x_1 \lt x_2 \le n \}$, $n \ge 8$, and let $\mathcal{G} = \{G_1, \ldots, G_n\}$ be a partition of $\mathcal{F}$ in $n$ parts.

For any $G_i \in \mathcal{G}$ and $X = \{x_1, x_2\} \in \mathcal{F}$, let $g(i, X) = \min_{1 \le j \le 2}|\{ Y \in G_i : Y \cap X = \{x_j\} \}|$. Let $h(X) = \sum_{i=1}^n g(i,X)$.

Is it possible to prove or disprove that for any $\mathcal{G}$ there always exists $Z \in \mathcal{F}$ such that $h(Z) \ge \Big\lfloor \frac{n-1}{2} \Big\rfloor$?

Motivation related to the union-closed sets conjecture: suppose we have a union-closed family $\mathcal{H}$ with $n$ sets, and the intersection of any $\Big\lfloor \frac{n+1}{2} \Big\rfloor$ sets in $\mathcal{H}$ is empty, so that $\mathcal{H}$ is a counterexample to the union-closed sets conjecture. Extending the reasoning in the answers to this question it is not difficult to show that if we build $\mathcal{F}$ from $\mathcal{H}$, with a bijection between $(\{x_1, x_2\}, G_i)$ and $(\{H_1, H_2\}, H_i)$, $H_1, H_2, H_i \in \mathcal{H}$, $H_1 \cup H_2 = H_i$, then it must be $h(X) \lt \Big\lfloor \frac{n-1}{2} \Big\rfloor$ for any $X = \{x_1, x_2\} \in \mathcal{F}$, otherwise we would have some $ p \not= q$ such that $H_p = H_q$. In other words, proving the statement in the question would imply the union-closed sets conjecture. But the converse is not true, therefore we have a strengthening of the conjecture.

EDIT 1

I had to modify the minimum $n$ in the first sentence from $n \ge 4$ into $n \ge 8$ because I found counterexamples for $n = 5$ and $n = 7$. However, from $8$ above, by taking $30000$ random partitions for each value of $n$, it seems that the statement in the question holds, with a margin increasing with $n$. Clearly, random samples do not prove anything. Maybe we could try to make an integer linear program model of the problem. Here is $\lfloor (n-1)/2 \rfloor$ (first line) compared with the minimum of the maximum value of $h(X)$ (second line) for $n \ge 3$ and $30000$ tests:

1,1,2,2,3,3,4,4,5,5,6,6,7,7,8, 8, 9, 9,10,10,11,11,12,12,13,13,14,14,15,15
0,1,1,2,2,3,4,4,5,5,6,7,8,8,9,10,10,11,12,12,13,13,14,15,15,15,17,17,18,18

EDIT 2

I have tried to solve an Integer Linear Program for the problem, with the following minimums for $3 \le n \le 7$: $0,1,1,1,1$. Unfortunately, while the solver takes a few seconds for $n = 7$, it gets out of memory after about $5$ hours for $n = 8$.

The question can be formulated in terms of graph edge labeling of a complete graph $K_n$ with up to $n$ labels.

Answer 1

A family of counterexamples was found at math.stackexchange using projective planes as I should have expected. I have copied the answer here for convenience:

Let $n = q^2+q+1$ and consider a projective plane of order $q$. It has $n$ points and $n$ lines such that every line contains $q+1$ points and every point lies on $q+1$ lines. Through every two points, there is a unique line, and every two lines intersect at a unique point.

If $\ell_1, \dots, \ell_n$ are the lines, let $G_i$ to be the set of all pairs of points on $\ell_i$, for $i=1, \dots, n$.

Now consider a pair $X = \{x_1, x_2\}$. There is a unique line $\ell_i$ containing $x_1, x_2$, and $q-1$ other points. For that line, we will have $g(i,X) = q-1$: there are $q-1$ pairs $\{x_1, x\} \in > G_i$ and $q-1$ pairs $\{x_2, x\} \in G_i$, where $x \ne x_1, x_2$.

Every other line $\ell_j$ contains at most one of $x_1$ or $x_2$. If $\ell_j$ does not contain $x_1$, then there are $0$ pairs of the form $\{x_1, x\}$ in $G_j$, and so $g(j,X) = 0$; similarly if $\ell_j$ does not contain $x_2$. Therefore $g(j,X) = 0$ for all $j \ne i$.

Therefore $h(X) = q-1$, and this is true for any $X$. As a result, the highest possible value of $h$ for this partition is far from $\lfloor \frac{n-1}{2}\rfloor$: it is closer to $\sqrt n$.