7.6 On a circle with a circumference of 1956, what is the minimum number of points that should be selected so that for each of these points, there is exactly 1 point at a distance of 1, and exactly 1 point at a distance of 2 (the distance between two points is measured along the arc)?
[Solution] Divide the circumference into 1956 segments using 1956 equally spaced points, then the distance between any two adjacent points is 1. Number these points in counterclockwise order as $1, 2, \cdots, 1956$. Then erase the points whose numbers are multiples of 3, leaving 1304 points, which clearly satisfy the requirements of the problem. Therefore, the minimum number of points required is no more than 1304.

On the other hand, let $S$ be any set of points that meets the requirements, and let $A \in S$. Then there exist points $B, C \in S$ such that the arc $\overparen{A B}$ has length 1 and the arc $\overparen{A C}$ has length 2. It is easy to prove that points $B$ and $C$ must lie on opposite sides of point $A$, otherwise it would lead to both $B$ and $C$ being at a distance of 1 from $A$, which is a contradiction. Without loss of generality, assume point $B$ is in front and point $C$ is behind. For point $B \in S$, there is already a point $A$ at a distance of 1 from it, so there must be another point $D \in S$ such that the arc $\overparen{D B}$ has length 2. Continuing this process, we find a sequence of points all belonging to $S$ and with adjacent points alternating distances of 1 and 2, until we reach a point behind $C$ that is at a distance of 1 from $C$. This results in a total of 1304 points in $S$. This means that any set of points that meets the requirements must have at least 1304 points.
In summary, the minimum number of points in a set that meets the requirements is 1304.