We partition $\{L_1,L_2,\dots,L_{100}\}$ into three sets. Let
$$\begin{align*} X &= \{L_n\mid n\equiv0\pmod{4}\}, \\ Y &= \{L_n\mid n\equiv1\pmod{4}\}, \\ Z &= \{L_n\mid n\equiv2,3\pmod{4}\}, \\ \end{align*}$$
from which $|X|=|Y|=25$ and $|Z|=50.$
Any two distinct lines can intersect at most once. To maximize the number of points of intersection, note that each point must be passed by exactly two lines. If three or more lines pass through the same point, then we can create more points of intersection by translating the lines.
We construct the sets one by one:

We construct all lines in set $X.$
Since the lines in set $X$ are parallel to each other, they have $0$ points of intersection.

We construct all lines in set $Y.$
The lines in set $Y$ have $1$ point of intersection with each other, namely $A.$
Moreover, each line in set $Y$ can intersect each line in set $X$ once. So, there are $25^2=625$ points of intersection.
Now, we have $\boldsymbol{1+625=626}$ additional points of intersection.

We construct all lines in set $Z.$
The lines in set $Z$ can have $\binom{50}{2}=1225$ points of intersection with each other.
Moreover, each line in set $Z$ can intersect each line in sets $X$ and $Y$ once. So, there are $50^2=2500$ points of intersection.
Now, we have $\boldsymbol{1225+2500=3725}$ additional points of intersection.

Together, the answer is $626+3725=\boxed{\textbf{(B) }4351}.$
~MRENTHUSIASM