Let's take 1996 lines on the plane such that any two of them intersect, but no three pass through the same point. The lines divide the plane into regions. What is the minimum and maximum number of angular regions among these?
For at least 3 and at most 1995, and generally for $n$ lines, the largest odd number that is not greater than $n$.

1. Let's call a point in the plane a vertex if it is the intersection of two given lines. A vertex is precisely the vertex of an (angle domain bounded by given lines) if it is the extreme vertex on both bounding half-lines, meaning one of the two half-lines it defines no longer contains any other vertices. A vertex cannot be the vertex of two angle domains, since the exclusion of multiple intersection points would only occur if two intersecting lines define the two angle domains; in that case, however, one of the lines does not contain any further vertices, or the other lines do not intersect it, contradicting the condition of the problem.
2. The convex hull of the (finite) set of vertices is at least a triangle, and since every vertex of this is an extreme vertex on the lines defining it, the number of angle domains created is at least 3.
3. Among the planar domains created by the lines, the unbounded ones are bounded by two half-lines (and possibly some segments), and each half-line is the boundary of exactly two unbounded domains. Thus, the number of unbounded domains is equal to the number of (empty inside) half-lines created, which is exactly $2n$, since two such half-lines are created on each line. Let's call two unbounded domains adjacent if they have a common bounding half-line, and opposite if their bounding half-lines lie on the same two lines. Taking a circle that contains all vertices inside it, and traversing the boundary of this circle, and listing the (unbounded) domains in the order of this traversal, adjacent domains follow each other, and the opposite ones are the $n$-th in the order from the other. Of two adjacent ones, at most one can be an angle domain, since the vertices of two adjacent angle domains are different, so one of them would be an interior point of the common bounding half-line. Thus, for odd $n$, we have obtained that the number of angle domains is at most $n$. Now let's assume that $n$ is even, and there are $n$ angle domains; then the angle domains stand in opposite pairs, which implies the existence of a common vertex, which is impossible. For even $n$, therefore, the number of angle domains is at most $n-1$.
4. If $n=3$, then clearly exactly 3 angle domains are created. Suppose we already have $n-1$ lines that define exactly 3 angle domains. Let's add the $n$-th line such that it intersects at least two of these angle domains in both of their bounding half-lines. Out of the infinite possibilities, only a finite number do not meet the condition of the intersection points, so it is achievable that the existing $n$ lines also satisfy the conditions of the problem. The $n$-th line added eliminates at least two of the originally angle domains (those whose both sides it intersects), and can create at most two new ones, since their vertices are extreme vertices on the new line, and there are only two of them. This proves that for every $n$, there exist $n$ lines that create exactly 3 angle domains.
5. Finally, we show that the upper limit is also achievable for every $n$. For odd $n$, connect each of the $n$ points on a circle to its $\left[\frac{n}{2}\right]$-th successor. The lines thus created intersect each other inside or on the boundary of the circle, so each of the $n$ points will be the vertex of an angle domain, and the number of lines is also $n$. For even $n$, perform the above construction with $(n-1)$. Take a diameter of the circle that passes through one of the vertices on the circumference, then rotate it slightly around the center of the circle so that the conditions given for the intersection points are met. The resulting diameter intersects exactly one of the original angle domains, and splits it into one angle domain and one unbounded domain that is not an angle domain; thus, the number of angle domains remains $n-1$, while the number of lines is $n$.

Péter Frenkel (Fazekas M. Főv. Gyak. Gimn. III. o.t.)