Given seven points on a circle, connect each pair of points. Find the minimum number of intersection points of these lines inside the circle.
We prove that these lines intersect at most three sets of three lines at a point inside the circle, and it is obvious that there cannot be four lines intersecting at a point.
First, select seven vertices of a regular octagon,
it is clear that there are three sets of three lines intersecting at a point (as shown in Figure 3).
Second, there cannot be more than three sets of three lines intersecting at a point. We call a point other than the intersection of three lines a "remaining point".
If there are two adjacent remaining points, assume without loss of generality that these seven points are $A_{1}, A_{2}, \cdots, A_{7}$, and $A_{1}$ and $A_{2}$ are both remaining points. Then, when $A_{1}$ is a remaining point, the intersection of three lines is the intersection of $A_{2} A_{5}$, $A_{3} A_{6}$, and $A_{4} A_{7}$. When $A_{2}$ is a remaining point, the intersection of three lines is the intersection of $A_{1} A_{5}$, $A_{3} A_{6}$, and $A_{4} A_{7}$. Both intersections of three lines are the intersection of $A_{3} A_{6}$ and $A_{4} A_{7}$, hence they coincide, which is a contradiction.
Therefore, remaining points cannot be adjacent.
Thus, among the seven points, there can be at most three remaining points, meaning there can be at most three sets of three lines intersecting at a point.
In summary, the minimum number of intersection points inside the circle is $\mathrm{C}_{7}^{4}-3(3-1)=29$.
(Zhang Lei, Northeast Yucai School, 110179)