Four. (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 $m$ groups, with two players competing in each group. In the next round, the players are regrouped to compete again, 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 tournament schedule where, after $n$ rounds, for any four players, they have either not played against each other at all or have played against each other at least twice in total.
For $m \geqslant 3, n_{\text {min }}=m-1$.
Assume that $t(t \leqslant m-2)$ rounds of matches have been played according to a feasible schedule.

Let $k$ be the largest positive integer such that there exist $A_{1}, A_{2}, \cdots, A_{k}$ players who have all played against each other. Then, $A_{1}$, in addition to playing against $A_{2}, \cdots, A_{k}$, has also played against $t-(k-1)=t-k+1$ 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
$$
\begin{array}{l}
2(t-k+1)+k=2 t-k+2 \\
\leqslant 2(m-2)=2 m-k-2 .
\end{array}
$$

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 these $k+2$ players. By the assumption about $k$, there exist $B_{i}$ and $B_{j}$ such that $B_{i}$ and $B_{j}$ have not played against each other, thus $A_{1}, A_{2}, B_{i}, B_{j}$ have played only one match among them. Therefore,
$$
n_{\text {min }} \geqslant m-1 \text {. }
$$

Next, we prove that $n_{\text {min }} \leqslant m-1$.
To do this, we need to show that there exists a possible schedule such that after $m-1$ rounds, for any four players, they have either not played against each other at all or have played at least two matches in total.
(1) $m$ is even.
Arrange the matches as follows:
For $1 \leqslant k \leqslant m-1$, in the $k$-th round:
If $i+j \equiv 2 k(\bmod m-1)(1 \leqslant i, j \leqslant m-1, i \neq j)$, then $i$ and $j$ are in the same group, and $i+m$ 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 next $m$ rounds, the groups are as follows:
$$
\begin{array}{l}
(1, m+1),(2, m+2), \cdots,(m, 2 m), \\
(1, m+2),(2, m+3), \cdots,(m, m+1), \\
\cdots \cdots \cdots
\end{array}
$$
$$
(1,2 m),(2, m+1), \cdots,(m, 2 m-1) \text {. }
$$

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, so any four players have played at least two matches in total.
Thus, $n_{\text {min }} \leqslant m-1$.
In conclusion, $n_{\text {min }}=m-1$.
(2) $m$ is odd.
Arrange the matches as follows:
For the first $m-1$ rounds:
$(1, m+2),(2, m+3), \cdots,(m, m+1)$, $(1, m+3),(2, m+4), \cdots,(m, m+2)$, ......
$(1,2 m),(2, m+1), \cdots,(m, 2 m-1)$.
For the next $m$ 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, and $i+m$ and $j+m$ play, and $k$ and $k+m$ play. This way, after $m-1$ rounds, any four players have either not played against each other at all or have played at least two matches in total.
Thus, $n_{\min } \leqslant m-1$.
In conclusion, $n_{\text {min }}=m-1$.