Filthy rich

I will give a generalised proof for $n \geq 3$ lines.

Define a based isosceles triangle to be a triangle formed by three of the lines, with two sides of equal length marked. Then the amount of money obtained by Calvin is equal to the number of based isosceles triangles.

For each ordered pair of distinct lines $(l_1, l_2)$ , there is at most one isosceles triangle with $l_1$ unmarked and $l_2$ marked. So there are at most $\lfloor \frac{n-1}{2} \rfloor$ based isosceles triangles with $l_1$ unmarked. Therefore, there are at most $n \times \lfloor \frac{n-1}{2} \rfloor$ based isosceles triangles altogether.

If $n$ is odd, Calvin can achieve this upper bound by drawing the lines to form the sides of a regular polygon with $n$ sides. If $n$ is even, Calvin can achieve this upper bound by drawing the lines to form all but one of the sides of a regular polygon with $n + 1$ sides.

Substituting $n=100$ , we get Calvin profits $4900.