2019 AMC 10A Problem 14

The values $0,1,3,4,5,$ and $6$ are attainable. Four parallel lines give $0$, four concurrent lines give $1$, three parallel lines cut by a fourth give $3$, three concurrent lines plus a fourth not through that point give $4$, three lines forming a triangle plus a fourth parallel to one side give $5$, and four lines in general position give $\binom42=6$. It remains to rule out $2$. Choose two nonparallel lines, meeting at $X$. If a third line also passes through $X$, then a fourth line not through $X$ intersects at least two of those three concurrent lines at two different new points; otherwise all four lines pass through $X$, giving only one point. If the third line does not pass through $X$, then to create only one new point it must be parallel to one of the first two lines. A fourth distinct line cannot pass through either existing intersection without meeting the parallel line at a new point, and if it passes through neither, it creates a new intersection immediately. Thus exactly two intersection points are impossible. Thus the possible values are $0,1,3,4,5,6$, whose sum is $19$. Thus, D is the correct answer.