Join two guests with an edge $u\sim v$ if and only if they are not friends with each other. Then $\sim$ is an equivalence relation (i.e. the graph can be decomposed into a union of pairwise disconnected cliques) and every guest has degree $n-5$. Let the cliques have sizes $c_1,...,c_r$. Then $4 = c_1 + ... + c_r - c_j \qquad \forall \quad j$, and each clique has size $n - 4$. Thus $(r-1)(n-4) = 4$ and $n \ge 5$. From here we should be able to bash out all cases...

Look at the prob on an [n x n] grid. Then you have to remove stepwise [n - 4] lines and rows (each). Only solutions: 5, 6, 8.