Three. (50 points)
Given a positive integer $m \geqslant 17$, there are $2 m$ players participating in a round-robin tournament. Each round, the $2 m$ players are divided into $m$ pairs, with each pair competing against each other. In the next round, the players are re-paired, and this continues for a total of $2 m-1$ rounds, ensuring that each player competes against each of the other $2 m-1$ players exactly once.

Find the smallest possible positive integer $n$ such that there exists a feasible schedule after a few rounds, where for any 4 players, they have either not competed against each other at all or have competed against each other at least twice.
For $m \geqslant 3$, the minimum possible $n_{\text {min }}=m-1$.
Suppose a feasible program has been carried out for $l$ rounds: $l \leqslant m-2$.
Let $k$ be the maximum positive integer such that there exist $k$ players who have all played against each other. Without loss of generality, let these $k$ players be $A_{1}, A_{2}, \cdots, A_{k}$. Then $A_{1}$, apart from having played against $A_{2}, A_{3}, \cdots, A_{k}$, has also played against $l-(k-1)=l-k+1$ other players. The same applies to $A_{2}$. Therefore, the number of players who have played against both $A_{1}$ and $A_{2}$ is at most $2(l-k+1)+k=2 l-k+2 \leqslant 2(m-2)-k+2=2 m-k-2$.

Thus, there are at least $k+2$ players who have not played against either $A_{1}$ or $A_{2}$. Let $B_{1}, B_{2}, \cdots, B_{k}+2$ be $k+2$ of these players. By the assumption about $k$, there exist $B_{i}, B_{j}$ who have not played against each other. Thus, among $A_{1}, A_{2}, B_{i}, B_{j}$, only one match has been played. Therefore, $n_{\text {min }} \geqslant m-1$.
Next, we prove that $n_{\min } \leqslant m-1$.
To do this, we need to show that there exists a feasible program such that after $m-1$ rounds, for any 4 players, they have either not played against each other at all or have played at least 2 matches in total.
1. If $m$ is even.
Arrange the matches as follows: for $1 \leqslant k \leqslant m-1$, in the $k$-th round:
If $i+j=2 k(\bmod m-1), 1 \leqslant i, j \leqslant m, i \neq j$, then $i$ and $j$ are in the same group. Also, $m$ and $j+m$ are in the same group: $k$ and $m$ are in the same group; $k+m$ and $2 m$ are in the same group. For the last $m$ rounds, the groups are arranged as follows:
$$
\begin{array}{c}
(1, m+1),(2, m+2), \cdots,(m, 2 m) \\
(1, m+2),(2, m+3), \cdots,(m, m+1) \\
\cdots \quad \cdots \quad \cdots \\
(1,2 m),(2, m+1), \cdots,(m, 2 m-1)
\end{array}
$$

After $m-1$ rounds, $1,2, \cdots, m$ have all played against each other, and $m+1, m+2, \cdots, 2 m$ have all played against each other. Therefore, any 4 players have played at least 2 matches in total.
Thus, $n_{\text {min }} \leqslant m-1$, so $n_{\min }=m-1$.
2. If $m$ is odd.
Arrange the matches as follows: for the first $m-1$ rounds:
$$
\begin{array}{c}
(1, m+2),(2, m+3), \cdots,(m, m+1) \\
(1, m+3),(2, m+4), \cdots,(m, m+2) \\
\cdots \quad \cdots \quad \cdots \\
(1,2 m),(2, m+1), \cdots,(m, 2 m-1)
\end{array}
$$

For the last $m$ rounds: $1 \leqslant k \leqslant m-1$
In the $m-1+k$-th round: if $i+j \equiv 2 k(\bmod m), 1 \leqslant i, j \leqslant m, i \neq j$, then $i$ and $j$ play; $i+m$ and $j+m$ play; $k$ and $k+m$ play. Thus, after $m-1$ rounds, any 4 players have either not played against each other at all or have played at least 2 matches in total. Therefore, $n_{\min } \leqslant m-1$, so $n_{\min }=m-1$.