The ninety-eighth problem: Given a positive integer $\mathrm{m} \geq 3$. There are $2 \mathrm{~m}$ players participating in a round-robin tournament. Each round, the $2 \mathrm{~m}$ players are divided into $\mathrm{m}$ 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 positive integer $n$ such that there exists a feasible tournament schedule where, after $n$ rounds, for any four players, they have either not competed against each other at all or have competed against each other at least twice in total.
The 98th question:
Solution: Use $2m$ points to represent these $2m$ people. If two people compete in a certain round, connect an edge between the corresponding two points. After $2m-1$ rounds of competition, a complete graph $K_{2m}$ of $2m$ order is eventually obtained. Clearly, each round of competition corresponds to a set of perfect matchings in the graph; after $n$ rounds of competition, a $2m$-order $n$-regular graph is obtained.
(1) We first prove: If there exists a $2m$-order $n$-regular graph $G(V, E)$, where among any four points, there are either no edges or at least 2 edges (i.e., there is no edge between any four points), then $n \geq m-1$.

At this point, for any $uv \in E$, any two points in $V \backslash N(\{u, v\})$ are connected (otherwise, if there exist two points $w_1, w_2$ in $V \backslash N(\{u, v\})$ that are not connected, then $u, v, w_1, w_2$ have only one edge $uv$), thus the induced subgraph of $V \backslash N(\{u, v\})$ is a complete subgraph of at least $2m - |N(\{u, v\})|$ order.

Let the maximum complete subgraph of $G$ be $G_0(v_0, E_0)$, and let $V_0 = \{v_1, v_2, \ldots, v_k\}$, then $|N(\{v_1, v_2\})| \leq k + 2(n - k + 1) = 2n - k + 2$. According to the above analysis, the induced subgraph of $V \backslash N(\{v_1, v_2\})$ is a complete subgraph with at least $2m - (2n - k + 2)$ points. By the maximality of $G_0$, we know:
$$
2m - (2n - k + 2) \leq k \Rightarrow n \geq m-1
$$
(2) We then prove: When $n = m-1$, there exists a competition schedule that meets the conditions.
Let the $2m$ people be $1, 2, \ldots, 2m$.
(1) When $m$ is even, arrange the competition as follows:
For $1 \leq k \leq m-1$, in the $k$-th round: If $i + j \equiv 2k \pmod{m-1} (1 \leq i, j \leq m-1, i \neq j)$, then $i$ and $j$ are in the same group; $i+m$ and $j+m$ are in the same group; $k$ and $m$ are in the same group; $k+m$ and $2m$ are in the same group. The last $m$ rounds are grouped as follows:
$(1, m+1), (2, m+2), \ldots, (m, 2m)$;
$(1, m+2), (2, m+3), \ldots, (m, m+1)$;
$(1, 2m), (2, m+1), \ldots, (m, 2m-1)$.
After $m-1$ rounds of competition, $1, 2, \ldots, m$ have all competed with each other, and $m+1, m+2, \ldots, 2m$ have all competed with each other, so any four players have competed at least twice in total.
(2) When $m$ is odd, arrange the competition as follows:
The first $m-1$ rounds:
$$
\begin{array}{l}
(1, m+2), (2, m+3), \ldots, (m, m+1) \\
(1, m+3), (2, m+4), \ldots, (m, m+2) \\
\cdots \cdots
\end{array}
$$
$(1, 2m), (2, m+1), \ldots, (m, 2m-1)$.
The last $m$ rounds: For any $1 \leq k \leq m$, in the $(m-1+k)$-th round, if $i + j \equiv 2k \pmod{m} (1 \leq i, j \leq m, i \neq j)$, then $i$ and $j$ are in the same group; $i+m$ and $j+m$ are in the same group; $k$ and $k+m$ are in the same group.
After $m-1$ rounds of competition, any four players have either not competed or have competed at least twice in total.
In summary, the minimum value of $n$ is $m-1$.
Note: When $m = 2k$ is even, it is easy to prove: There exists a scheme such that after $m-1$ rounds of competition, $1, 2, \ldots, m$ have all competed with each other, and $m+1, m+2, \ldots, 2m$ have all competed with each other.

For this, it is only necessary to prove: $K_m$ can be divided into $m-1$ sets of perfect matchings (1, 2, \ldots, m have all competed with each other, corresponding to a $K_m$; $m+1, m+2, \ldots, 2m$ correspond to another $K_m$. It is only necessary to prove that a $K_m$ can be divided into $m-1$ sets of perfect matchings, and the other is similar).
In fact, arrange the $2k$ vertices of $K_m$ as follows:
In the graph, each row of two vertices is adjacent, clearly forming a 1-factor of $K_m$.
Then move one position in the direction of the arrows in the graph, and another 1-factor of $K_m$ is obtained. Repeat this process $2k-1$ times, and $2k-1$ edge-disjoint 1-factors of $K_m$ are obtained, whose union is exactly $K_m$.
This is the construction of the first part of (1). The construction of the second part of (2) is similar.