Assume otherwise, then there is a violating subset for each potential added element. This amounts to at least $n-29$ violating subsets. But each subset includes two elements in the 29 element nice subset, so there are at most ${29 \choose 2}$ of these. Therefore $n \leq {30 \choose 2}$. Claim: There is such an example for $n = {30 \choose 2}$. Construction: Consider the set of pairs of elements of $\{1, 2, \ldots, 30\}$. Create a three-element subset for every triple $\{a, b\}, \{b, c\}, \{c, a\}$. Obviously these satisfy the condition. Now the 29-element nice subset $\{\{1, 2\}, \{1, 3\}, \ldots \{1, 30\}\}$ cannot have any other pairs added for obvious reasons. hence the minimum value of $n$ is ${30 \choose 2}+1 = 436$

Inconsistent wrote: Assume otherwise, then there is a violating subset for each potential added element. This amounts to at least $n-29$ violating subsets. But each subset includes two elements in the 29 element nice subset, so there are at most ${29 \choose 2}$ of these. Therefore $n \leq {30 \choose 2}$. Claim: There is such an example for $n = {30 \choose 2}$. Construction: Consider the set of pairs of elements of $\{1, 2, \ldots, 30\}$. Create a three-element subset for every triple $\{a, b\}, \{b, c\}, \{c, a\}$. Obviously these satisfy the condition. Now the 29-element nice subset $\{\{1, 2\}, \{1, 3\}, \ldots \{1, 30\}\}$ cannot have any other pairs added for obvious reasons. hence the minimum value of $n$ is ${30 \choose 2}+1 = 436$ Could you please tell the motivation of selecting the construction? I think the hard and main part of this problem is to find such construction.