222 Find the smallest natural number $n$, such that for any $n$ points taken on a circle, among the central angles of the arcs connecting any two points, at least 2005 angles are not greater than $120^{\circ}$.
Parse First, explain that 90 points do not work. On a pair of diametrically opposite points on the circle, take 45 points near each, at this time there are only $2 \cdot \mathrm{C}_{45}^{2}=1980$ angles not exceeding $120^{\circ}$.
Next, explain that $n=91$ has this property.
Consider the 91 points on the circumference of the circle as vertices of a graph, and stipulate that if the central angle between two points exceeds $120^{\circ}$, a line is drawn between these two points, thus forming the graph $G$. Let the number of edges be $x$.
From the construction of the graph, it is known: the graph contains no triangles.
Suppose $A_{1}$ and $A_{2}$ are connected, then among the remaining 89 points, there is no point that is connected to both $A_{1}$ and $A_{2}$. Let $d\left(A_{i}\right)$ denote the degree of $A_{i}$, then
$\left(d\left(A_{1}\right)-1\right)+\left(d\left(A_{2}\right)-1\right) \leqslant 89$. That is, $d\left(A_{1}\right)+d\left(A_{2}\right) \leqslant 91$.
$\sum_{i=1}^{m} d\left(A_{i}\right)=2 x$. For each pair of vertices corresponding to an edge, $d\left(A_{i}\right)+d\left(A_{j}\right) \leqslant 91$. Summing up, we get:
$\sum_{i=1}^{91} d^{2}\left(A_{i}\right) \leqslant 91 x$. By the Cauchy inequality, we know: $91 x \geqslant \sum_{i=1}^{91} d^{2}\left(\dot{A}_{i}\right) \geqslant \frac{1}{91}\left(\sum_{i=1}^{91} d\left(A_{i}\right)\right)^{2}$ $=\frac{4 x^{2}}{91}$.
Therefore, $x \leqslant \frac{91^{2}}{4}$. Since $x \in \mathbf{N}$, we have $x \leqslant 2070$.
Thus, the number of point pairs with a central angle not exceeding $120^{\circ}$ is $\geqslant \mathrm{C}_{91}^{2}-2070>2005$.
In summary, the minimum value of $n$ is 91.
Comment From the above discussion, it can be seen that the 2005 in this problem can be any integer between 1981 and 2025.