Let $A = \{a_1, a_2, \dots, a_k\}$ and $B = \{b_1, b_2, \dots, b_k\}$ be the two friendly groups. For arbitrary group $C$ from $A$ denote by $S_C$ the group of all people from $B$ each of which is an enemy of at least one person from $C$. If $|C| > |S_C|$ then $C \cup (B \setminus S_C)$ is a friendly group having more than $k$ members, a contradiction.

Therefore the sets $S_{\{a_i\}}$ satisfy the Hall's condition for system of distinct representatives. Hence, we may assume that $a_i$ and $b_i$ are enemies for all $i = 1, 2, \dots, k$.

This means that every friendly group of $k$ members include one person from every pair $(a_i, b_i)$ for $i = 1, 2, \dots, k$.

Suppose the enemies of $a_1$ are $b_1, \dots, b_t$. For a friendly group $S$ such that $a_1 \in S$ we have $b_1, \dots, b_t \notin S$. Hence, $a_2, \dots, a_t \in S$. From every of the remaining $k-t$ pairs $(a_j, b_j)$, $j > t$ we have to choose one of $a_j$ and $b_j$ to be an element of $S$. Therefore there are at most $2^{k-t}$ such group.

For a friendly group $S$ with $k$ members such that $a_1 \notin S$ we have $b_1 \in S$. Since from each of the remaining $k-1$ pairs $(a_j, b_j)$, $j > 1$ we have to choose one of $a_j$ and $b_j$ to be an element of $S$ we find that there are at most $2^{k-1}$ such groups.

Therefore, the total number of friendly groups of $k$ members is at most $2^{k-1} + 2^{k-t}$.