Let the smallest positive integer $k$ satisfying the condition of the problem be denoted $r(m, n)$. We shall show that
$$r(m, n) = 2(m + n) - \min\{m, n\} - 1$$
Observe that, by symmetry, $r(m, n) = r(n, m)$. Therefore it suffices to consider the case where $m \geq n$, and to prove that
$$r(m, n) = 2m + n - 1. \quad[1 \text{ mark }] \tag{1}$$
First we prove that
$$r(m, n) \geq 2m + n - 1$$
by an example. Call a group of $k$ people, every two of whom are mutually acquainted, a $k$-clique. Consider a set of $2m + n - 2$ people consisting of a $(2m - 1)$-clique together with an additional $n - 1$ people none of whom know anyone else. (Call such people isolated.) Then there are not $2m$ people forming $m$ mutually acquainted pairs, and there also are not $2n$ people forming $n$ mutually unacquainted pairs. Thus $r(m, n) \geq (2m - 1) + (n - 1) + 1 = 2m + n - 1$ by the definition of $r(m, n)$. [1 mark]

To establish (1), we need to prove that $r(m, n) \leq 2m + n - 1$. To do this, we now show that
$$r(m, n) \leq r(m - 1, n - 1) + 3 \quad \text{for all } m \geq n \geq 2. \tag{2}$$
Let $G$ be a group of $t = r(m - 1, n - 1) + 3$ people. Notice that
$$t \geq 2(m - 1) + (n - 1) - 1 + 3 = 2m + n - 1 \geq 2m \geq 2n$$
If $G$ is a $t$-clique, then $G$ contains $2m$ people forming $m$ mutually acquainted pairs, and if $G$ has only isolated people, then $G$ contains $2n$ people forming $n$ mutually unacquainted pairs. Otherwise, there are three people in $G$, say $a, b$ and $c$, such that $a, b$ are acquainted but $a, c$ are not. Now consider the group $A$ obtained by removing $a, b$ and $c$ from $G$. $A$ has $t - 3 = r(m - 1, n - 1)$ people, so by the definition of $r(m - 1, n - 1)$, $A$ either contains $2(m - 1)$ people forming $m - 1$ mutually acquainted pairs, or else contains $2(n - 1)$ people forming $n - 1$ mutually unacquainted pairs. In the former case, we add the acquainted pair $a, b$ to $A$ to form $m$ mutually acquainted pairs in $G$. In the latter case, we add the unacquainted pair $a, c$ to $A$ to form $n$ mutually unacquainted pairs in $G$. This proves (2). [3 marks]

Trivially, $r(s, 1) = 2s$ for all $s$ [1 mark], so $r(m, n) \leq 2m + n - 1$ holds whenever $n = 1$. Proceeding by induction on $n$, by (2) we obtain
$$r(m, n) \leq r(m - 1, n - 1) + 3 \leq 2(m - 1) + (n - 1) - 1 + 3 = 2m + n - 1,$$
which completes the proof. [1 mark]