Given two positive integers $m$ and $n$, find the smallest positive integer $k$ such that among any $k$ people, either there are $2m$ of them who form $m$ pairs of mutually acquainted people or there are $2n$ of them forming $n$ pairs of mutually unacquainted people.
1. **Define the problem in terms of graph theory:**
   - We are given two positive integers \( m \) and \( n \).
   - We need to find the smallest positive integer \( k \) such that among any \( k \) people, either there are \( 2m \) of them who form \( m \) pairs of mutually acquainted people or there are \( 2n \) of them forming \( n \) pairs of mutually unacquainted people.

2. **State the answer and the claim:**
   - The answer is \( 2\max(m, n) + \min(m, n) - 1 \).
   - We claim that this is the minimal \( k \).

3. **Construct a counterexample to show minimality:**
   - Consider a complete graph with \( 2m + n - 2 \) vertices.
   - Create a clique of \( 2m-1 \) people who all know each other.
   - Add \( n-1 \) people who don't know anyone else.
   - The largest matchings in this graph have sizes \( m-1 \) and \( n-1 \) respectively.

4. **Define the function \( f(m, n) \):**
   - Let \( f(m, n) \) be the smallest \( k \) such that the conditions are satisfied.

5. **Assume \( m > n \) without loss of generality:**
   - Since \( f(m, n) = f(n, m) \), we can assume \( m > n \).

6. **Color the edges and use a procedural recoloring method:**
   - Call acquainted people red and not acquainted people blue.
   - Assume we have a counterexample to the hypothesis.
   - Recolor edges until the largest red matching has size exactly \( m-1 \).

7. **Describe the recoloring process:**
   - Let the vertices be \( I_1, I_2, \ldots, I_{n+1} \) and \( J_1, J_2, \ldots, J_{2m-2} \).
   - Assume there is a matching between \( J_1 \) and \( J_2 \), \( J_3 \) and \( J_4 \), and so on, and this is maximal.
   - Start with vertex \( V \) as \( I_1 \).
   - Observe that \( VJ_1 \) and \( I_2J_2 \) are not both red, otherwise we would have a larger matching.
   - One of these edges is blue. If \( V \) is part of the blue edge, let \( V = I_2 \). Otherwise, keep \( V \)'s value.
   - Repeat this process with \( VJ_3 \) and \( I_3J_3 \).

8. **Handle the case when \( m = n \):**
   - After applying the procedure \( n-1 = m-1 \) times, two vertices are left.
   - If the edge between them is red, we have an \( m \)-red matching.
   - If the edge between them is blue, we have an \( n \)-blue matching.

9. **Conclude the proof:**
   - Hence, we are done and the answer is \( 2\max(m, n) + \min(m, n) - 1 \).

The final answer is \( \boxed{ 2\max(m, n) + \min(m, n) - 1 } \).