Solution:

Let the lines all have slope $\frac{p}{q}$ where $p$ and $q$ are relatively prime. Without loss of generality, let this slope be positive. Consider the set of points that consists of the point of $S$ with the smallest coordinates on each individual line in the set $L$. Consider a point $(x, y)$ in this, because there is no other point in $S$ on this line with smaller coordinates, either $x \leq q$ or $y \leq p$. Additionally, since each line passes through at least two points in $S$, we need $x+q \leq 100$ and $y+p \leq 100$.

The shape of this set of points will then be either a rectangle from $(1,1)$ to $(100-q, 100-p)$ with the rectangle from $(q+1, p+1)$ to $(100-q, 100-p)$ removed, or if $100-q<q+1$ or $100-p<p+1$, just the initial rectangle. This leads us to two formulas for the number of lines,
$$N= \begin{cases}(100-p)(100-q)-(100-2 p)(100-2 q) & p, q<50 \\ (100-p)(100-q) & \text{ otherwise }\end{cases}$$
In the first case, we need to minimize the quantity
$$\begin{aligned} (100-p)(100-q)-(100-2 p)(100-2 q) & =100(p+q)-3 p q \\ & =\frac{10000}{3}-3\left(q-\frac{100}{3}\right)\left(p-\frac{100}{3}\right) \end{aligned}$$
if one of $p, q$ is above $100 / 3$ and the other is below it, we would want to maximize how far these two are from $100 / 3$. The case $(p, q)=(49,1)$ will be the optimal case since all other combinations will have $p, q$'s closer to $100 / 3$, this gives us $4853$ cases.

In the second case, we need to minimize $p$ and $q$ while keeping at least one above $50$ and them relatively prime. From here we need only check $(p, q)=(50,1)$ since for all other cases, we can reduce either $p$ or $q$ to increase the count. This case gives a maximum of $4950$.