Question 1 Find the smallest positive integer $n$ such that: for any $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ taken on the circumference of $\odot O$, among the $\mathrm{C}_{n}^{2}$ angles $\angle A_{i} O A_{j}(1 \leqslant i<j \leqslant n)$, at least 2007 are not greater than $120^{\circ}$.
【Analysis】Let $A B$ be a diameter of $\odot O$. Near point $A$ (rotating $O A$ clockwise by less than $30^{\circ}$ to get $O A^{\prime}$, and counterclockwise by less than $30^{\circ}$ to get $O A^{\prime \prime}$, on the arc $\overparen{A^{\prime} A^{\prime \prime}}$), place $p$ points. Near point $B$ (rotating $O B$ clockwise by less than $30^{\circ}$ to get $O B^{\prime}$, and counterclockwise by less than $30^{\circ}$ to get $O B^{\prime \prime}$, on the arc $\widetilde{B^{\prime} B^{\prime \prime}}$), place $q$ points. Select one point $X$ near $A$ and one point $Y$ near $B$, then $\angle X O Y > 120^{\circ}$.

If two points $X_{1}, X_{2}$ are selected near $A$ or two points $Y_{1}, Y_{2}$ are selected near $B$, then $\angle X_{1} O X_{2}, \angle Y_{1} O Y_{2}$ are both no greater than $120^{\circ}$.

In fact, if a line is drawn between any two points $X, Y$ that satisfy $\angle X O Y > 120^{\circ}$, it forms a complete bipartite graph.
Let the number of angles no greater than $120^{\circ}$ be $S$.
First, make a preliminary estimate of the value of $n$.
When $p = q = 45$,
$S = \mathrm{C}_{45}^{2} + \mathrm{C}_{45}^{2} = 19802007$.
If $n = 90$, by equation (1), it is known that there is at least one selection method that does not satisfy $\mathrm{C}_{n}^{2}$ angles $\angle A_{i} O A_{j} (1 \leqslant i < j \leqslant n)$ all being greater than $120^{\circ} (1 \leqslant i < j \leqslant n)$.
Therefore, $n = 91$ is the smallest positive integer that satisfies the condition.