On a finite set $S$, the topology is uniquely determined by the closures of singletons: A subset of $S$ is closed iff it contains the closure of all its elements.
Given a topology, we can define
$$\tag1a\le b\iff \overline{\{a\}}\subseteq \overline{\{b\}}$$
(or equivalently, $a\in \overline{\{b\}}$).
This need not be a partial order because we can have $a\le b$ and $b\le a$ with $a\ne b$. But it is a reflexive and transitive relation on $S$.
On the other hand, from a a reflexive and transitive relation $\le$ on $S$, we can define a topology by declaring
$$\tag2 A\text{ closed}\iff\forall a\in A\colon \forall x\in S\colon x\le a\to x\in A.$$
One verifies that the union or intersection of two closed sets is closed and trivially $\emptyset$ and $S$ are closed.
Moreover, the associations topology $\leftrightarrow$ reflexive transitive relation in $(1)$ and $(2)$ are inverse of each other, hence
On a finite set, there are as many topologies as there are reflexive transitive binary relations.
The relation to partial orders is that a reflexive transitive binary relation on $S$ is the same as a partial order on a partition of $S$. In particular, if the counts of topologies and partial orders are $T_n$ and $P_n$, respectively, we have
$$ T_n=P_n+{n\choose 2}P_{n-1}+\left({n\choose 3}+\frac12{n\choose 2}{n-2\choose 2}\right)P_{n-2}+\ldots+2^{n-1}P_2+P_1.$$