I think n must be 15.

can you post solution?

I think the answer is 15,too Here is my solution,and please tell me if it's wrong For $i=1,2,..,15$ ,let $x_i=|\{j\in[1,n]|i \in A_j\}|$ Then we have the following result: 1)$\sum_{i=1}^{15}x_i=7n$ 2)$\sum_{i=1}^{15}\binom{x_i}{2} \leq 3 \binom{n}{2}$ From 2),we get $\sum x_i^2-\sum x_i \leq 3n(n-1)$ Then aplying AM-GM and 1),we can easily get $n \geq 15$ Now we will prove that $n=15$ satisfies the hypothesis,and we just have to check the 3 rd condition . First,we notice that there does not exist 4 subsets that have 3 common elements (which is implied from the hypothesis) So we just consider the following cases: (*)Case 1:There exists unique $i,j,k$ (asumming they are 1,2,3)such that $|A_1 \cap A_2 \cap A_3|=\{1,2,3\}$. Then the remaining 4 elements of each subset will get all the value in the set {4,5,6,...,15} (notice that 3+3x4=15) We count the number of distiinct subset {x,y,z} where $x,y,z \in A_i$ The total is $3\binom{7}{3}-3+12\binom{7}{3}=522>\binom{15}{3}$ So the condition 3 is satisfied. (*)Case 2:There exists only 2 subsets that have 3 common elements. Like case 1,We count the number of distinct subset {x,y,z} where $x,y,z \in A_i$,the total is $2\binom{7}{3}-2+13\binom{7}{3}=523>\binom{15}{3}$ (*)Case 3:There does not exists indices i,j such that $|A_i \cap A_j|=3$ Like case 1 and 2,the total number of triples is $15\binom{7}{3}>\binom{15}{3}$ So we are done .

KDS wrote: From 2),we get $\sum x_i^2-\sum x_i \leq 3n(n-1)$ Then aplying AM-GM and 1),we can easily get $n \geq 15$ Can anyone explain this step please? I get using AM-GM and (1) $\frac{49n^2}{15}-7n = (\sum_{i=1}^{15}x_i)^2/15 - \sum_{i=1}^{15}x_i \leq 3n^2-3n \implies n\leq 15?$

Another way of bounding $n$: For all 3 element subsets $M$ of $S$, let $d(M)$ be the number of $A_i$ containing M. Let a, b be the number of 3 element subsets $M$ of $S$ such that $d(M)=3,2$ respectively. Since $S\backslash M$ has 12 elements, $d(M)\le 4$. Note that the number of $(M,A_i)$ such that $M$ is contained in $A_i$ is $$n\binom{7}{3} = 2a+b+\binom{15}{3} \hspace{2cm} (1).$$Suppose next $\forall i \in S, i$ is contained in $x_i$ subsets, it follows that $\sum x_i = 7n.$ Each $A_i\cap A_j$ has size $\leq 2$ unless their intersection is some $M$ (of size 3), so $$n\binom{7}{3}- \binom{15}{3} + (n\binom{7}{3}- \binom{15}{3})/2 +2\binom{n}{2} \stackrel{(1)}{\ge} 3a+b+2\binom{n}{2} = \sum \binom{d(M)}{2} + 2\binom{n}{2}$$$$\ge \text{ number of $(i,\{j,k\})$ pairs with $i\in A_j\cap A_k$} = \sum \binom{x_i}{2}$$$$\ge \frac{1}{2}((\sum_{i=1}^{n}x_i)^2/n - \sum_{i=1}^{n}x_i) =\frac{1}{2}(\frac{49n^2}{n}-7n),$$so in shorter terms $$3(n\binom{7}{3}- \binom{15}{3})/2 +2\binom{n}{2} \ge \frac{1}{2}(\frac{49n^2}{n}-7n)$$. It follows from the inequality above that $n \geq 15$. A third way: Consider $F$, the set of sets $A_i$ containing $x$ for an arbitrary $x\in S$. Since any two sets in $F$ after the exclusion of $x$ have intersection at most 2 but by (III) their union has size 14, it follows that each $x\in S$ is contained in $\ge 7$ sets $A_i$. Thus $3\binom{n}{2} \geq 15\binom{7}{2} \implies n\geq 15.$