SOLVED
The cochromatic number of $G$, denoted by $\zeta(G)$, is the minimum number of colours needed to colour the vertices of $G$ such that each colour class induces either a complete graph or empty graph. Let $z(n)$ be the maximum value of $\zeta(G)$ over all graphs $G$ with $n$ vertices.

Determine $z(n)$ for small values of $z(n)$. In particular is it true that $z(12)=4$?

A question of Erdős and Gimbel, who knew that $4\leq z(12)\leq 5$ and $5\leq z(15)\leq 6$. The equality $z(12)=4$ would follow from proving that if $G$ is a graph on $12$ vertices such that both $G$ and its complement are $K_4$-free then either $\chi(G)\leq 4$ or $\chi(G^c)\leq 4$.

In fact there do exist such graphs - Bhavik Mehta found computationally that there is exactly one (up to taking the complement) graph on $12$ vertices such that both $G$ and its complement are $K_4$-free with chromatic number $\geq 5$. This graph was explicitly checked to have cochromatic number $4$, and hence this proves that indeed $z(12)=4$.

The values of $z(n)$ are now known for $1\leq n\leq 19$: $1,1,2,2,3,3,3,3,4,4,4,4,5,5,5,6,6,6,6.$ (The only significant difficulty here is proving $z(12)=4$ - the others follow from easy inductive arguments and the facts that $R(3)=6$ and $R(4)=18$.) It is unknown whether $z(20)=6$ or $7$.

Gimbel [Gi86] has shown that $z(n) \asymp \frac{n}{\log n}$.

Additional thanks to: Bhavik Mehta