Problem: For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$
$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$

Solution: It is possible to obtain $0$ $1$ $3$ $4$ $5$ , and $6$ points of intersection, as demonstrated in the following figures: [asy] unitsize(2cm); real d = 2.5; draw((-1,.6)--(1,.6),Arrows); draw((-1,.2)--(1,.2),Arrows); draw((-1,-.2)--(1,-.2),Arrows); draw((-1,-.6)--(1,-.6),Arrows);  draw((-1+d,0)--(1+d,0),Arrows); draw((0+d,1)--(0+d,-1),Arrows); draw(dir(45)+(d,0)--dir(45+180)+(d,0),Arrows); draw(dir(135)+(d,0)--dir(135+180)+(d,0),Arrows); dot((0+d,0));  draw((-1+2*d,sqrt(3)/3)--(1+2*d,sqrt(3)/3),Arrows); draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows); draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows); draw((-1+2*d,-sqrt(3)/6)--(1+2*d,-sqrt(3)/6),Arrows); dot((0+2*d,sqrt(3)/3)); dot((-1/2+2*d,-sqrt(3)/6)); dot((1/2+2*d,-sqrt(3)/6));  draw((-1/3,1-d)--(-1/3,-1-d),Arrows);  draw((1/3,1-d)--(1/3,-1-d),Arrows); draw((-1,-1/3-d)--(1,-1/3-d),Arrows); draw((-1,1/3-d)--(1,1/3-d),Arrows); dot((1/3,1/3-d)); dot((-1/3,1/3-d)); dot((1/3,-1/3-d)); dot((-1/3,-1/3-d));  draw((-1+d,sqrt(3)/12-d)--(1+d,sqrt(3)/12-d),Arrows); draw((-1/4-1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((1/4+1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((-1+d,-sqrt(3)/6-d)--(1+d,-sqrt(3)/6-d),Arrows); dot((0+d,sqrt(3)/3-d)); dot((-1/2+d,-sqrt(3)/6-d)); dot((1/2+d,-sqrt(3)/6-d)); dot((-1/4+d,sqrt(3)/12-d)); dot((1/4+d,sqrt(3)/12-d));  draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw(dir(30)+(2*d,-d)--dir(30+180)+(2*d,-d),Arrows); draw(dir(150)+(2*d,-d)--dir(-30)+(2*d,-d),Arrows); dot((0+2*d,0-d)); dot((0+2*d,sqrt(3)/3-d)); dot((-1/2+2*d,-sqrt(3)/6-d)); dot((1/2+2*d,-sqrt(3)/6-d)); dot((-1/4+2*d,sqrt(3)/12-d)); dot((1/4+2*d,sqrt(3)/12-d)); [/asy]
It is clear that the maximum number of possible intersections is ${4 \choose 2} = 6$ , since each pair of lines can intersect at most once.  We now prove that it is impossible to obtain two intersections.
We proceed by contradiction.  Assume a configuration of four lines exists such that there exist only two intersection points.  Let these intersection points be $A$ and $B$ .  Consider two cases:
Case 1 :  No line passes through both $A$ and $B$
Then, since an intersection point is obtained by an intersection between at least two lines, two lines pass through each of $A$ and $B$ .  Then, since there can be no additional intersections, the 2 lines that pass through $A$ cant intersect the 2 lines that pass through $B$ , and so 2 lines passing through $A$ must be parallel to 2 lines passing through $B$ . Then the two lines passing through $B$ are parallel to each other by transitivity of parallelism, so they coincide, contradiction.
Case 2 :  There is a line passing through $A$ and $B$
Then there must be a line $l_a$ passing through $A$ , and a line $l_b$ passing through $B$ .  These lines must be parallel.  The fourth line $l$ must pass through either $A$ or $B$ .  Without loss of generality, suppose $l$ passes through $A$ .  Then since $l$ and $l_a$ cannot coincide, they cannot be parallel.  Then $l$ and $l_b$ cannot be parallel either, so they intersect, contradiction.
All possibilities have been exhausted, and thus we can conclude that two intersections is impossible.  Our answer is given by the sum $0+1+3+4+5+6=\boxed{19}$

Answer: D 19