Four. (50 points) On a plane, there are 100 different points and $n$ different lines $l_{1}, l_{2}, \cdots, l_{n}$, and the number of points that line $l_{k}$ passes through is denoted as $a_{k}$. If $a_{1}+a_{2}+\cdots+a_{n}=250$, find the minimum possible value of $n$.


First, prove that $n \geqslant 21$.
Let the number of points that lie exactly on $k(k=0,1, \cdots, n)$ lines be $x_{k}$.
Then $\sum_{k=0}^{n} x_{k}=100, \sum_{k=0}^{n} k x_{k}=\sum_{k=1}^{n} a_{k}=250$.
Thus, $\sum_{k=2}^{n} x_{k} \leqslant 100$,
$\sum_{k=2}^{n}(k-1) x_{k}=150+x_{0} \geqslant 150$.
Considering the number of intersection points of the lines, we have
$\sum_{k=2}^{n} \mathrm{C}_{k}^{2} x_{k} \leqslant \mathrm{C}_{n}^{2}$.
For any integer $k$, we have
$(k-2)(k-3) \geqslant 0$
$\Rightarrow \frac{k(k-1)}{2} \geqslant 2 k-3$.
Thus, for $k \geqslant 2$, $\mathrm{C}_{k}^{2} \geqslant 2 k-3$.
Hence, $\mathrm{C}_{n}^{2} \geqslant \sum_{k=2}^{n} \mathrm{C}_{k}^{2} x_{k}$
$\geqslant 2 \sum_{k=2}^{n}(k-1) x_{k}-\sum_{k=2}^{n} x_{k}$
$\geqslant 2 \times 150-100=200$
$\Rightarrow n \geqslant 21$.
Next, prove that when $n=21$, such points and lines exist.
When 100 points include 3 points not on any line, 1 point on only one line, 39 points as intersections of two lines, and 57 points as intersections of three lines,
$$
a_{1}+a_{2}+\cdots+a_{21}=1+39 \times 2+57 \times 3=250,
$$

satisfying the conditions.
It remains to construct 21 lines that intersect pairwise, with 57 sets of three lines intersecting at a common point.
Consider the unit circle $x^{2}+y^{2}=1$, and let
$$
P(\alpha)=(\cos \alpha, \sin \alpha),
$$

and the line $l(\alpha)=P(\alpha) P(\pi-2 \alpha)$, where, when $\alpha=\frac{\pi}{3}, \pi$,
$\frac{5 \pi}{3}$, $l(\alpha)$ is a tangent to the circle.
Take 20 lines $l_{i}=l\left(\frac{i \pi}{10}\right)(i=1,2, \cdots, 20)$, and it is known that these 20 lines intersect pairwise.
The lines $l(\alpha) 、 l(\beta) 、 l(\gamma)$ intersect at a common point if and only if
$\alpha+\beta+\gamma \equiv 0(\bmod 2 \pi)$.
Considering directed arcs, we have
$\overparen{P(\alpha) P(\beta)}+\overparen{P(\gamma) P(\pi-2 \beta)}$
$\equiv \beta-\alpha+\pi-2 \beta-\gamma \equiv \pi(\bmod 2 \pi)$,
$l(\beta) \perp P(\alpha) P(\gamma)$.
Similarly, $l(\alpha) \perp P(\beta) P(\gamma)$.
Thus, $l(\alpha) 、 l(\beta) 、 l(\gamma)$ intersect at the orthocenter of $\triangle P(\alpha) P(\beta) P(\gamma)$.
If $\gamma^{\prime}>\gamma$, then $\pi-2 \gamma^{\prime}<\pi-2 \gamma$.
Hence, $l\left(\gamma^{\prime}\right)$ does not pass through the intersection point of $l(\alpha) 、 l(\beta) 、 l(\gamma)$.
Similarly, for $\gamma^{\prime}<\gamma$.
Therefore, these 20 lines do not have any four lines intersecting at a common point.
Note that, $\alpha \checkmark \beta$ has $20 \times 19$ different choices, with 19 choices each satisfying
$\alpha+2 \beta \equiv 0(\bmod 2 \pi)$ or $2 \alpha+\beta \equiv 0(\bmod 2 \pi)$, disregarding the order of $\alpha, \beta, \gamma$, there are $\frac{20 \times 19-2 \times 19}{3!}=57$ sets of three lines intersecting at a common point.

At this point, there must exist a line $l_{21}$ that intersects with $l_{1}, l_{2}, \cdots, l_{20}$, and does not have any three lines intersecting at a common point. This satisfies the requirement for the number of intersection points.