Number of Door Spaces on a 4-point Set

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Question from S. Morris's Topology Without Tears on Door spaces: (see related issues here and here)

A "Door Space" is a topology $(X, \tau)$in which every subset of $X$ is either open or closed (or both).

For $X=\{a,b,c,d\}$, which topologies are Door Spaces?

The question is a little vague -- I'm not sure if he means which of the specific topologies already introduced in his text are Door Spaces, or which of the 33 inequivalent topologies on a 4-point set are Door Spaces.

If the former, the problem is pretty easy; I don't need help there.

If the latter (which of the 33), a brute force approach will work. But is there a better way to count Door Spaces of an n-point set?

For a 4-point set, there are 14 proper subsets that make up 7 partitions into pairs. So any 4-point Door Space must contain at least 7 subsets (in addition to $X$ and $\emptyset$) in its topology in order to account for every subset as open or closed.

From here, I can just look up the explicit list of 4-point topologies and check/count the members that have more 7 or more elements.

Are there more criteria I can use to reduce the number to check/count?

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I think it should be pretty easy to enumerate all (non-homeomorphic) 33 topologies on a 4 point set (I did it myself once), it's a bit tedious, but OK. Then check which ones are door spaces.

Counting the open sets is one easy filtering criterion, I agree.

Also there is going to be a discrete part (all $x$ such that $\{x\}$ is open), and then you can concentrate the counting on the remainder part (all closed singletons). So such spaces are an open discrete subspace unioned with a closed relatively discrete subspace, it seems to me. This quite limits the options.