Let $ S=\{1,2,\ldots,9 \}$ and $ \mathcal{T}= \{A \subset S \ ; \ |A|=5 \}$. Find the minimum value of $n$ such that for any $ \mathcal{U} \subset \mathcal{T} $ with $|\mathcal{U}|=n$ there exist two sets $A,B \in \mathcal{U}$ so that $ |A \cap B|=4$.
Let $\mathcal{F}=\{A_1,A_2,\ldots,A_k\} \subset \mathcal{T}$ be a family of $5$-element subsets of $S$ such that: $$|A\cap B|\leq 3\;\;\;\;\;\;\forall A,B \in \mathcal{F}, A\neq B$$ Now we are interested in $k_{\max}$.
Make a following bipartite graph. Connect $A_i$ with a $4$-element subset $B\subset S$ if and only if $B\subset A_i$.
Then the degree of any $B$ is at most $1$, while the degree of each $A_i$ is $ {5\choose 4}=5$. So we have $1\cdot {9\choose 4}\geq 5 \cdot k$, and so $k\leq 25$. Thus the partial answer is $n_{min}\leq 26$.
Now I don't know how to find a configuration for $n=26$. If I understand we are searching for Steiner system $S(4,5,9)$?
Continuing the thread in my comment above, I did a computer search for a counterexample using this Python implementation of Algorithm X. I found no solutions. You can run the program yourself here. To help convince you I coded everything correctly, I used the same algorithm to find $S(5,6,12)$ by brute force. You can play with the code to see it find/not find $S(k-1,k,n)$ for other values of $k,n$, and verify this agrees with the (non)existence of Steiner $k$-tuple systems for your chosen $n$.
In short, I think $n_{\text{min}}\le 25$.