The bound $h(n) \ll n^{2/3}$ is achievable by the following twisted cubic construction: Let $X=\mathbb{F}_p^3$ ($p\geq5$) and $Y_{ij}=\{(t,t^2+i,t^3+j): t \in \mathbb{F}_p\}$. The sets $Y_{ij}$ partition $X$ and $Y_{ij} \cup Y_{kl}$ never contains any 4-AP.
If there were a 4-AP, $\{a,b,c,d\}$, there must be two elements belonging to $Y_{ij}$ and two belonging to $Y_{kl}$, since none of $Y_{ij}$ contains a 3-AP. We may assume $\{a,b\} \in Y_{ij}$ and $\{c,d\} \in Y_{kl}$. In this case, the vectors $a-b$ and $c-d$ would be parallel, but a twisted cubic does not contain any pair of secants that are parallel.
We can "project" $\mathbb{F}_p^3$ on $\mathbb{Z}$ by sending $(a,b,c)$ (taking values in $0 ... p-1$) to $4ap^2+2bp+c$. In this projection, if a subset of $\mathbb{F}_p^3$ does not contain a 4-AP, the image would not contain a 4-AP either. So we can project the $Y_{ij}$s on $\mathbb Z$ without creating any 4-APs. By translating the projected $Y_{ij}$s by $xp^2+yp+z$$4xp^3+2yp^2+zp$ $(x,y,z \in \{0,1\})$ and renaming the variables, we can partition $0 ... 8p^3-1$ by $8p^2$ sets.
Thus the bound $h(n) \ll n^{2/3}$ is achievable.