In the interior of a circle centred at $O$ consider the $1200$ points $A_1,A_2,\ldots ,A_{1200}$, where for every $i,j$ with $1\le i\le j\le 1200$, the points $O,A_i$ and $A_j$ are not collinear. Prove that there exist the points $M$ and $N$ on the circle, with $\angle MON=30^{\circ}$, such that in the interior of the angle $\angle MON$ lie exactly $100$ points.