2. Let $n(n \geqslant 4)$ be a positive integer. $n$ players each play a table tennis match against every other player (each match has a winner and a loser). Find the minimum value of $n$ such that after all the matches, there always exists an ordered quartet $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$, satisfying that when $1 \leqslant i<j \leqslant 4$, player $A_{i}$ defeats player $A_{j}$.
(Supplied by He Yixie)
2. First, prove: when $n=8$, there always exists an ordered quartet that satisfies the problem's conditions.

Since 8 players have played a total of $\mathrm{C}_{8}^{2}=28$ matches, there must be at least one player who has won at least $\left\lceil\frac{28}{8}\right\rceil=4$ matches (where $\left\lceil x \right\rceil$ denotes the smallest integer not less than the real number $x$).
Assume player $A_{1}$ has defeated $a_{1}, a_{2}, a_{3}, a_{4}$.
Among players $a_{1}, a_{2}, a_{3}, a_{4}$, a total of six matches have been played, so at least one player must have won at least $\left\lceil\frac{6}{4}\right\rceil=2$ matches. Assume player $a_{1}$ has defeated players $a_{2}$ and $a_{3}$, and assume player $a_{2}$ has defeated player $a_{3}$. Then, set players $A_{2}, A_{3}, A_{4}$ to be $a_{1}, a_{2}, a_{3}$, respectively. Therefore, the ordered quartet $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$ satisfies the conditions.

Next, prove: when $n \leqslant 7$, an ordered quartet that satisfies the conditions does not necessarily exist.
It suffices to disprove the case for $n=7$.
Label the 7 players as $b_{1}, b_{2}, \cdots, b_{7}$, and agree that $b_{7+k}=b_{k}$. Construct the following scenario:

For $i=1,2, \cdots, 7$, let player $b_{i}$ defeat players $b_{i+1}, b_{i+2}, b_{i+4}$, but lose to players $b_{i+3}, b_{i+5}, b_{i+6}$.
This precisely determines the outcome of every match.
Assume there exists an ordered quartet $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$ that meets the conditions.
Since player $A_{1}$ must be one of the $b_{1}$, players $A_{2}, A_{3}, A_{4}$ can only be a permutation of $b_{i+1}, b_{i+2}, b_{i+4}$. However, because player $b_{i+1}$ defeats player $b_{i+2}$, player $b_{i+2}$ defeats player $b_{i+4}$, and player $b_{i+4}$ defeats player $b_{i+1}$, none of the players $b_{i+1}, b_{i+2}, b_{i+4}$ can serve as player $A_{4}$, leading to a contradiction.
In conclusion, the smallest value of $n$ that satisfies the conditions is 8.