Let $G(V,E)$ be the simple undirected graph where $V = \{v_1, v_2, \dots, v_{2n}\}$ and $E = \{(v_i,v_j) | v_i \text{ and } v_j \text{ are friends}\}$. Now, it is given that every $v_i$ has more friends than enemies. This implies that any $v_i$ has atleast $n$ friends and atmost $n-1$ enemies.
Mathematically, $\forall_i \ deg(v_i) \ge n$, in other words, every vertex of graph $G$ has degree more than $n$.
If total guests are $2$, then both must be friends and we will have a hamiltonian path.
If total guests are $2n\ge4$ and degree of every vertex is atleast $n$, then graph is Hamiltonian. ($\textbf{Dirac's Theoram}$)
Presence of hamiltonian cycle shows that there exist a sequence of $2n$ vertices where every two consecutive vertices are connected implying that both are friends.
Hence, we can seat them around the round table while making sure that everyone has two friends as their neighbours.
