We divide the circle into three equal sectors of $120^\circ$ such that none of the points belong to their border, with the exception of the center of the circle. If the center of the circle is one of the points, then we consider that it belongs only to one of the sectors. This is possible because the number of the points is finite and the possible selections for the distribution of the circle into three equal sectors are infinite.

Let $A, B$ belong to the same sector which is defined by the radii $OK$ and $O\Lambda$. Then, if $OA$, $OB$ intersect the circle at $A'$, $B'$ and $A'OB = \omega < 60^\circ$, then we get $AB \leq A'B' = 2R \sin(\omega) < 2R \sin 60^\circ = 2 \sqrt{3}/2 = \sqrt{3}$. Equality is not possible because the points do not belong to the border of the sector. Hence two arbitrary points of the same sector have distance less than $\sqrt{3}$.

Let, in the three sectors, belong $x$, $y$, $z$ points, respectively. Then $x + y + z = 111$ and the segments with length less than $\sqrt{3}$ are at least
$$\binom{x}{2} + \binom{y}{2} + \binom{z}{2} = \frac{x(x-1) + y(y-1) + z(z-1)}{2} = \frac{x^2 + y^2 + z^2 - 111}{2}.$$
From Cauchy-Schwarz inequality: $x^2 + y^2 + z^2 \geq \frac{(x + y + z)^2}{3} = \frac{111^2}{3}$ and
$$\binom{x}{2} + \binom{y}{2} + \binom{z}{2} \geq \frac{111^2}{3} - 111 = 1998.$$