If $P$ is a topological property, then a space $(X, \tau)$ is said to be minimal $P$ (respectively, maximal) if $(X, \tau)$ has property $P$ but no topology on $X$ which is strictly smaller (respectively, strictly larger ) than $\tau$ has $P$.
The spaces are called $KC$-spaces in which every compact subset is closed.
I know that every minimal $KC$-topological space is compact.
Is every minimal $KC$-topological space, maximal compact?
Yes, this is true. We will prove a stronger result:
$$KC+\text{compact}\Longrightarrow \text{maximal compact}$$
proof: let $(X,\tau)$ be a compact KC-topology. Assume it's not maximal compact, i.e. there exists a compact topology $\tau'$ on X such that $\tau'$ is strictly finer than $\tau$. Then there exists a $\tau'$-closed set $A$ which is not $\tau$-closed. Since $(X,\tau')$ is compact, $A$ is $\tau'$-compact. A compact set in a finer topology is also compact in a coarser topology. Hence $A$ is $\tau$-compact. As $(X,\tau)$ is KC, $A$ is $\tau$-closed, contrary to the choice of $A$. Hence the result follows.
Since you already knew that every minimal KC space is compact (a proof can be found here), we are done.
The converse of the result in the quote box also holds.