A regular 1976-sided polygon has the midpoints of all its sides and diagonals marked. Of the points thus obtained, what is the maximum number that can lie on a circle?
I. solution. It is easily seen that there is a circle passing through 1976 points. We will show that more points cannot lie on a circle.

Consider a circle $k$ containing certain midpoints, and the diagonals of the polygon for which these are precisely the midpoints (if this is not unique, take one such diagonal). If from each vertex of the polygon, at most 2 of the above diagonals emanate, then the number of diagonals is at most 1976, because each diagonal was counted at two vertices. Thus, the number of midpoints on the circle is also at most 1976.

If the above case does not hold, then there is a vertex $A$ of the polygon from which at least 3 diagonals have midpoints on the circle $k_1$. These three midpoints are clearly distinct, thus they determine $k$. On the other hand, if we shrink the circumcircle of the polygon by half from point $A$, the resulting circle $k_1$ contains the midpoints of the diagonals emanating from $A$. Thus, $k_1$ must be identical to $k$. On $k$, there are certainly all the midpoints of the diagonals emanating from $A$ (including the center $O$ of the circumcircle of the polygon), which is a total of 1975 points.

Now we will show that no other midpoint can be on $k_1$.

Let $d$ be a diagonal and $F$ its midpoint on $k$. Since the segment connecting the center of the circle to the midpoint of the chord is perpendicular to the chord, $O F \perp d$, and since $A O$ is a diameter in circle $k$, by Thales' theorem, $d$ must pass through the other endpoint of the diameter of $k$, which is $A$. Thus, indeed, only the midpoints of the diagonals passing through $A$ can be on $k$.

Thus, at most 1976 of the marked points can lie on a circle.

II. solution. Draw all the diagonals (including sides) from one vertex of the polygon. Among these, one is the diameter of the circumcircle $k$ of the polygon, and the others are pairwise symmetric with respect to this diameter. Thus, $\frac{1975-1}{2}=987$ diagonals of different lengths are obtained. Each of these diagonals is at a different distance from the center $O$ of the circumcircle of the polygon. Draw a circle around $O$ with each of these distances. Thus, we have 987 concentric circles. We know that if we draw chords of equal length in a circle, the locus of their midpoints is a circle concentric with the original circle, whose radius is the distance of the chord from the center. By selecting any of the 987 circles, this circle thus contains the midpoints of all the diagonals of equal length. Diagonals of equal length can be drawn 1976 times in the polygon, so there are 1976 midpoints on each concentric circle.

Now consider a circle different from the above. This circle can intersect each of the 987 circles in at most 2 points, that is, it contains at most 1974 midpoints, possibly including the $O$ point. This totals 1975 midpoints.

We have thus seen again that at most 1976 of the midpoints can lie on a circle.