#### Solution

We consider all segments with endpoints at the marked points. Since no three of the marked 64 points lie on a single straight line, there will be $64\times63$/2 = 2016 $>$ 1981 such segments. From problem 116573 it follows that among these segments there will be at least one pair of parallel points. A quadrilateral with vertices at the ends of these segments is the desired one $($it cannot be a parallelogram, since there is no parallelogram with vertices at the vertices of the regular 1981-gon$)$.