What is the smallest number of points that can be chosen on a circle of length 1956 so that for each of these points there is exactly one chosen point at a distance of 1 and exactly one at a distance of 2 (distances are measured along the circumference)?

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Answer: 1304. Let $A$ be one of the selected points, $B$ and $C$ be the selected points, which are at distances 1 and 2 from it, respectively. The arrangement in the order $A B C$ is impossible, since in this case, for point $B$, there are two selected points at a distance of 1. Therefore, the points are arranged in the following order: $C \quad A B$ (or $B A \quad C$). Let $D$ be the point at a distance of 1 from $C$. The arrangement $C D A B$ is obviously impossible. Therefore, the arrangement is as follows: $D C \quad A B$. Let, further, $E$ be the point at a distance of 1 from $B$. It is arranged as follows: $D C \quad A B \quad E$. Continuing these considerations, we will see that the circle of length 1956 will be divided into 1956/3 $=652$ arcs of length 3 (the ends of these arcs are points $A, C, E, \ldots$). On each arc, there is one point. In total, we get $2 \cdot 652=1304$ points. All these points must necessarily be present.