Is every Kolmogorov ($T_0$) space either Fréchet ($T_1$) or sober or both, or is it possible to be Kolmogorov but neither of those two? If the latter, can a counterexample be defined?
A sober space is one in which every irreducible closed set is the closure of exactly one singleton. A closed set is irreducible if it's not a sum of two proper closed subsets.
For $x\in\Bbb R$ let $U_x=(\leftarrow,x)=\{y\in\Bbb R:y<x\}$, and let $\tau=\{\varnothing,\Bbb R\}\cup\{U_x:x\in\Bbb R\}$; $\tau$ is a $T_0$ topology on $\Bbb R$ that is not $T_1$. The non-empty closed subsets of the space are $\Bbb R$ and the sets $\Bbb R\setminus U_x=[x,\to)$ for $x\in\Bbb R$. Each of these is irreducible. Each $\Bbb R\setminus U_x$ has $x$ as its unique generic point, but $\Bbb R$ itself has no generic point: for each $x\in\Bbb R$, $\operatorname{cl}\{x\}=\Bbb R\setminus U_x$.
Replace $\Bbb R$ by $\Bbb Q$ and let $\tau$ be the topology generated by the sets $U_x=\{y\in\Bbb Q:y<x\}$ for $x\in\Bbb Q$, and you get an even better example: if $\alpha$ is an irrational, $\{x\in\Bbb Q:x>\alpha\}$ is closed and irreducible and has no generic point.