I might make a mistake because i rarely can solve a combinatoric question.(China ) We have $(n-1)n/2$ winner and $n$ playerso if the ratio $\frac{n-1}{2}$ is greater than $3$ we can find a player who wins against at least $4$ players . But its easy to see that among $4$ players we can choose $a_2,a_3,a_4$ so we can choose that player as $a_1$. So $n=8$ holds conditions. For $n=7$ (i hope ) we can find an example such that every player win against $3$ player and in all groups of this $3$ players we cant choose such $a_i$ . So i think answer is $8$.

How can I prove that if n = 8 then there always exist such ordered four people group ? Thank you.

The answer is $\boxed{8}$. We first prove that it works. Take the obvious directed graph interpretation. Note that the total out-degree of this graph is $\tbinom{8}{2}=28$ so there exists a vertex $a$, which has out-arrow to $b,c,d,e$. Consider a directed graph generated by $b,c,d,e$. This graph has total out-degree $\tbinom{4}{2}=6$ so there exists a vertex (WLOG it is $b$) which has out-arrow to at least two vertices (WLOG they are $c,d$). If $\overrightarrow{cd}$ is a directed edge. Then $(a,b,c,d)$ satisfies the problem's condition. Otherwise $\overrightarrow{dc}$ is a directed edge which implies $(a,b,d,c)$ satisfies the problems' condition. Hence we can conclude the first part. Now we have to give counter example for $n=7$. Work on $\mathbb{Z}/7\mathbb{Z}$ and label the vertices $1,2,3,4,5,6,7$ respectively. Let $\overrightarrow{ab}$ is a directed edge if and only if $a-b\in\{1,2,4\}$. Since each vertex has in-degree and out-degree $3$, the directed $K_4$ must be in form $\{a,a+1,a+2,a+4\}$. But it's easy to see that we have a directed cycle $a+1\to a+2\to a+4$, which is contradiction. Hence we are done.