= Solution
Use the uniform <Frankl-Wilson theorem>: if $p$ is a <prime number>, $L\subseteq\mathbb F_p$ has $s$ distinct residues, $s\leq\min\{r,n-r\}$, and an $r$-<uniform set family> $\mathcal F$ has distinct-member intersection sizes in $L$ modulo $p$ while $r\bmod p\notin L$, then
$$
|\mathcal F|\leq\binom ns.
$$
Here is a <polynomial method in combinatorics> proof. For each $A\in\mathcal F$, form the <intersection polynomial>
$$
f_A(x)=\prod_{\lambda\in L}\left(\sum_{i\in A}x_i-\lambda\right)
$$
over $\mathbb F_p$, and use <multilinear reduction on the Boolean cube> so its degree is at most $s$. At the <characteristic vector> of $B\in\mathcal F$, this evaluates to zero for $A\ne B$ and is nonzero for $A=B$. Consequently these evaluation functions are <linearly independent>.
The <low-degree evaluation rank on a uniform layer> bounds their space by $\binom ns$. To justify that step without division by a possibly zero residue, form the integer incidence matrix with rows $T\subseteq[n]$, $|T|\leq s$, columns $r$-sets $B$, and entries $\mathbf1_{T\subseteq B}$. Over the <rational numbers>, a row with $|T|=j$ equals $\binom{r-j}{s-j}^{-1}$ times the sum of the rows of all $s$-sets containing $T$. Thus its rational <matrix rank> is at most $\binom ns$. Every larger minor is an integer <determinant> equal to zero and stays zero modulo $p$, so the same upper bound holds over $\mathbb F_p$. This proves the theorem.
For the forbidden midpoint problem, split $\mathcal A$ according to whether its members contain coordinate 1. Leave the containing half unchanged; complement every member of the avoiding half. Both resulting families consist of $2p$-sets containing 1, and complementation preserves their internal intersection sizes because
$$
|A^c\cap B^c|=4p-|A|-|B|+|A\cap B|=|A\cap B|.
$$
Within either transformed family, distinct intersections lie between 1 and $2p-1$ and are not $p$. Their residues are therefore in $L=\{1,\ldots,p-1\}$, whereas the set-size residue is $2p\equiv0$. Apply the <Frankl-Wilson theorem> to each half and add:
$$
\boxed{|\mathcal A|\leq2\binom{4p}{p-1}}.
$$
\b[The factor 2 comes from splitting into two families.] The unsplit family can contain disjoint sets, whose intersection residue zero equals the set-size residue and prevents this direct application. Removing the shared coordinate after splitting also gives the stronger <complement splitting for forbidden midpoint intersections> bound $2\binom{4p-1}{p-1}$.
For the construction, fix a $(p+1)$-set $B$ and let $R=[4p]\setminus B$, of size $3p-1$. The <fixed-core construction avoiding midpoint intersections> is
$$
\mathcal F=\{B\cup X:X\in R^{(p-1)}\},\qquad
\mathcal A=\mathcal F\cup\{A^c:A\in\mathcal F\}.
$$
Two members of $\mathcal F$ intersect in at least $p+1$ points; the same is true for two complements. An intersection across the two halves has size $|X\setminus Y|\leq p-1$. No intersection is $p$. The halves are disjoint because one contains all of $B$ and the other none, so
$$
\boxed{|\mathcal A|=2\binom{3p-1}{p-1}}.
$$
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