Given positive integer $ m \geq 17$, $ 2m$ contestants participate in a circular competition. In each round, we devide the $ 2m$ contestants into $ m$ groups, and the two contestants in one group play against each other. The groups are re-divided in the next round. The contestants compete for $ 2m\minus{}1$ rounds so that each contestant has played a game with all the $ 2m\minus{}1$ players. Find the least possible positive integer $ n$, so that there exists a valid competition and after $ n$ rounds, for any $ 4$ contestants, non of them has played with the others or there have been at least $ 2$ games played within those $ 4$.
1. **Constructing the competition schedule for even \( m \):**
   - Label the contestants as \( A_1, A_2, \ldots, A_m, B_1, B_2, \ldots, B_m \).
   - For \( 1 \le i < m \), on day \( i \), pair \( A_j \) with \( A_{i + j} \) for each \( 1 \le j \le m \), where indices are modulo \( m \).
   - On day \( m \), pair \( A_j \) with \( B_j \) for \( 1 \le j \le m \).
   - For the last \( m-1 \) days, schedule so that the \( A_j \)'s play each other once and the \( B_j \)'s play each other once.

2. **Constructing the competition schedule for odd \( m \):**
   - Label the contestants in the same way as for even \( m \).
   - Schedule games in the same way for the first \( m-1 \) days.
   - For the \( (m-1+i) \)-th day, for \( 1 \le i \le m \), pair \( A_i \) with \( B_i \).
   - Pair the remaining players so that \( A_x \) plays \( A_{2i - x} \) and \( B_x \) plays \( B_{2i - x} \) for \( x \neq i \), where indices are again modulo \( m \).

3. **Proving that \( n = m-1 \) works:**
   - Suppose \( n \le m-2 \) worked. Consider the graph where two people are connected if they have played in the first \( n \) days. This graph is \( n \)-regular.
   - Let \( S \) be the largest set of people, none of whom have played each other. Suppose \( |S| = k \). By Turan's Theorem, \( k \ge 3 \).
   - If anyone else plays at most \( k-2 \) of the people in \( S \), we can find a 4-tuple with exactly one match played among them. Hence, assume everyone else plays at least \( k-1 \) of the people in \( S \).
   - This implies that, because each of the people in \( S \) plays \( n \) others, there are at least \( \frac{nk}{k-1} \) other people. Hence \( k + \frac{nk}{k-1} \ge 2m \).
   - Simplifying, this is equivalent to \( k^2 - (2m+1-n)k + 2n \ge 0 \).

4. **Analyzing the inequality:**
   - As \( n \le m-2 \), this implies \( k \ge 2m-n \).
   - Since there are at most \( n \) people not in \( S \), we have \( |S| = 2m-n \) and everybody in \( S \) must play all of the people not in \( S \).
   - This means that anyone not in \( S \) has played each of the \( 2m-n > n \) people in \( S \), which is a contradiction.

5. **Conclusion:**
   - Therefore, \( n \le m-2 \) does not work, and the answer is \( \boxed{m-1} \).