Now crossposted at math.stackexchange.
Consider a union-closed family $\mathcal{F} = \{A_1, \dotsc ,A_n\}$ of $n$ finite sets, $n$ odd, $n \ge 3$, $A_i \neq \emptyset$, $i=1,\dotsc,n$.
Let $r=\frac{n+1}{2}$. We have that:
$$\bigcup_{1 \le i_1 \lt \ldots \lt i_r \le n} A_{i_1} \cap \ldots \cap A_{i_r } = \bigcap_{1 \le i_1 \lt \ldots \lt i_r \le n} A_{i_1} \cup \ldots \cup A_{i_r} \tag{1}\label{1}$$
This is because the LHS contains all elements that are in at least $r$ of the $A_i$ and the RHS contains all elements that are in at least $n-r+1$ of the $A_i$ (because for each element that does not belong to the RHS we can find $r$ $A_i$ that do not contain it and thus it is in maximum $n-r$ of the $A_i$) and note that in our case $r=n-r+1$.
The RHS is the intersection of $\binom{n}{\frac{n+1}{2}}$ union expressions. However, we can simplify it removing all terms equal to $U(\mathcal{F})$ or which are supersets of another term, keeping only one union expression for each possible resulting set, to get:
$$\bigcap_{1 \le i_1 \lt \ldots \lt i_r \le n} A_{i_1} \cup \ldots \cup A_{i_r} = \bigcap_{k=1}^{h} A_{i_{k,1}} \cup \dotsb \cup A_{i_{k,r}} = \bigcap_{k=1}^{h} B_k \tag{2}\label{2}$$
Where $h \le n-1$ and $B_k = A_{i_{k,1}} \cup \dotsb \cup A_{i_{k,r}}$. After that, if there is no set $A_{i_{k,1}},\ldots,A_{i_{k,r}}$ in the expression equal to $B_k$, we replace one set with it, otherwise we leave the expression like it is. In any case we can rewrite it as:
$$B_k = A_{j_{k,1}} \cup \dotsb \cup A_{j_{k,r-1}} \cup B_k \tag{3}\label{3}$$
Now suppose that there is some $B_t = A_{j_{u,v}}$ with $t \not= u$. That would imply $B_t \subseteq B_u$ and then $\eqref{2}$ not minimal. Therefore $B_t \not= A_{j_{u,v}}$ and for what said before also $B_t \not= A_{j_{t,z}}$. This means that $A_{i_{k,1}},\ldots,A_{i_{k,r-1}}$ are chosen among $n-h-1$ sets ($-h$ is to remove the $B_k$ and $-1$ for $U(\mathcal{F})$).
Now suppose that $\bigcap_{k=1}^{h} B_k = \emptyset$.
Suppose also that there is one $A_l$ appearing in $h-1$ of the $\eqref{3}$ expressions, WLOG in $B_1, \ldots ,B_{h-1}$. Then $A_l \cap A_{i_{h,m}} = \emptyset$, $m=1,\ldots,r-1$ and $A_l \cap B_h = \emptyset$, therefore it is easy to see that the unions $A_l \cup A_{i_{h,m}}$, $m=1,\ldots,r-1$ and $A_l \cup B_h$ must be all different and different also from $A_l$, $A_{i_{h,m}}, m=1,\ldots,r-1$, and $B_h$. We would then have a total of $1 \times r+1+r=n+2 \gt n$ sets, absurd.
Therefore all $A_l$ must appear in maximum $h-2$ of the $\eqref{3}$ expressions.
As said above $A_{i_{k,1}},\ldots,A_{i_{k,r-1}}$ are $r-1=\frac{n-1}{2}$ sets chosen among $n-h-1$ sets, with $1 \le k \le h$. None of those sets can appear in more than $h-2$ union expressions. Therefore, thinking to a matrix where each column represents the same set and each row one of the union expressions, we can say that if $\bigcap_{k=1}^{h} B_k = \emptyset$ then:
$$h(n-h-1) \ge \frac{n-1}{2}h + 2(n-1-h) \label{4}\tag{4}$$
Inequality $\eqref{4}$ has no solution for $n$ odd, $n \lt 25$. It implies also that $h \le \frac{n-1}{2}$.
I am curious to see how looks a union-closed family such that all $A_l$ appear in maximum $h-2$ of the $\eqref{3}$ expressions, but I wasn't able to find an example. Someone can help?
I precise that in the example that I search it shouldn't be possible to replace some $B_k$ with a different expression, in a way that there is a set appearing $h-1$ or $h$ times.
A possible idea is starting from this example where $h=4$ but there is no set appearing in all the $4$ expressions, but there are sets appearing in $h-1=3$ expressions. However I can't see how to build an example as required by the question.
