$35 \cdot 34$ Lines $L_{1}, L_{2}, \cdots, L_{100}$ are distinct, and all lines $L_{4 n}$ are parallel to each other ($n$ is a positive integer); all lines $L_{4 n-3}$ pass through a known point $A$ ($n$ is a positive integer). The maximum number of intersection points in the set $\left\{L_{1}, L_{2}, \cdots, L_{100}\right\}$ is 
(A) 4350.
(B) 4351.
(C) 4900.
(D) 4901.
(E) 9851.
(27th American High School Mathematics Examination, 1976)
[Solution] 100 lines intersecting at most have $C_{100}^{2}=\frac{100 \cdot 99}{2}=4950$ intersection points.
However, lines $L_{4}, L_{8}, \cdots, L_{100}$ are parallel, thus losing $C_{25}^{2}=300$ intersection points.
Moreover, lines $L_{1}, L_{5}, \cdots, L_{97}$ intersect only at point $A$, so another $C_{25}^{2}-1=299$ intersection points are lost.
$\therefore$ The maximum number of intersection points is $\quad 4950-300-299=4351$. Therefore, the answer is $(B)$.