Solution:

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 $2$. Suppose the only intersection points are $X$ and $Y$. If no line passes through both, then the lines through $X$ must all be parallel to the lines through $Y$ to avoid new intersections, which is impossible because two distinct lines through $X$ are not parallel. If one line passes through both $X$ and $Y$, then any other line through $X$ and any other line through $Y$ must be parallel; the fourth line still creates an additional intersection unless it is parallel to both, which cannot cover both points.

Thus the possible values are $0,1,3,4,5,6$, whose sum is $19$. Thus, D is the correct answer.