Children disliking in a circle

Consider the children as vertices of a graph, where an edge connects two children who like each other. Then the problem is reduced to finding a Hamiltonian cycle in this graph.

By Dirac's theorem, all $n$-vertex graphs whose minimum degree is at least $\left\lceil\frac n2\right\rceil$ are Hamiltonian. This translates to a maximum $k$ of $$n-1-\left\lceil\frac n2\right\rceil=\left\lfloor\frac{n-2}2\right\rfloor$$ For example, if six children are such that each dislikes at most two others, we can always seat them around the table. If each dislikes three others, and the pattern of likes is as follows:

  A F
 /| |\
B | | E
 \| |/
  C D

then we cannot sit them around the table.

To show the sharpness of the bound for $k$ shown above, divide $n-1$ vertices into two sets $A$ and $B$ as equally as possible (so $|A|=\left\lceil\frac{n-1}2\right\rceil$ and $|B|=\left\lfloor\frac{n-1}2\right\rfloor$). Construct two graphs $K_{|A|}$ and $K_{|B|}$ and link the one remaining vertex to all the other vertices. The resulting graph has a maximal $k$ one more than the bound, but is non-Hamiltonian.