2019 AMC 12A Problem 8

The maximum is $\binom{4}{2} = 6$ when the lines are in general position

Enumerate configurations using parallel classes and concurrences; check which counts from $0$ to $6$ are achievable

The values $0$ and $1$ occur when all four lines are parallel or all four are concurrent. Three parallel lines crossed by a fourth give $3$ points. Three concurrent lines together with a fourth line not through their common point give $4.$ One parallel pair with no three concurrent gives $5,$ and four lines in general position give $6.$

The value $2$ is impossible. Once two lines meet at $P,$ a third line not through $P$ creates at least one new intersection; a fourth distinct line must then create another new point. If every remaining line passes through $P,$ there is only one intersection point instead. Thus the achievable values are $0,1,3,4,5,6,$ whose sum is $19.$

Thus, the correct answer is D.