12. (16 points) 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 $120^{\circ}$.
12. First, when $n=90$, as shown in Figure 7, let $AB$ be the diameter of $\odot O$. Take 45 points near points $A$ and $B$, respectively. At this time, there are only $2 \mathrm{C}_{45}^{2}=45 \times 44=1980$ angles that do not exceed $120^{\circ}$. Therefore, $n=90$ does not satisfy the problem's requirements.
Second, when $n=91$, we will prove that there are at least 2007 angles that do not exceed $120^{\circ}$.

For 91 points $A_{1}, A_{2}, \cdots, A_{91}$ on the circumference, if $\angle A_{i} O A_{j}>120^{\circ}$, then connect $A_{i} A_{j}$, thus forming a graph $G$. Suppose graph $G$ has $e$ edges.

When $\angle A_{i} O A_{j}>120^{\circ}, \angle A_{j} O A_{k}>120^{\circ}$, then $\angle A_{i} O A_{k} \leq 120^{\circ}$. Therefore, if $e \geqslant 1$, without loss of generality, assume $A_{1}$ and $A_{2}$ are connected by an edge. Since the graph has no triangles, for any point $A_{i} (3 \leqslant i \leqslant 91)$, it can be connected to at most one of $A_{1}$ or $A_{2}$. Thus,
$$
d\left(A_{1}\right)+d\left(A_{2}\right) \leqslant 89+2=91,
$$
where $d(A)$ denotes the number of edges from $A$.
$$
\text{Also, } d\left(A_{1}\right)+d\left(A_{2}\right)+\cdots+d\left(A_{91}\right)=2 e \text{, }
$$
and for every pair of vertices $A_{i}, A_{j}$ connected by an edge in graph $G$, we have $d\left(A_{i}\right)+d\left(A_{j}\right) \leqslant 91$.
Thus, summing over all edges, we get
$$
\left(d\left(A_{1}\right)\right)^{2}+\left(d\left(A_{2}\right)\right)^{2}+\cdots+\left(d\left(A_{91}\right)\right)^{2}
$$
$\leqslant 91 e$.
By the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
91\left[\left(d\left(A_{1}\right)\right)^{2}+\left(d\left(A_{2}\right)\right)^{2}+\cdots+\left(d\left(A_{91}\right)\right)^{2}\right] \\
\geqslant\left[d\left(A_{1}\right)+d\left(A_{2}\right) \cdots+d\left(A_{91}\right)\right]^{2}=4 e^{2} . \\
\text{Thus, } \frac{4 e^{2}}{91} \leqslant\left(d\left(A_{1}\right)\right)^{2}+\left(d\left(A_{2}\right)\right)^{2}+\cdots+\left(d\left(A_{91}\right)\right)^{2} \\
\leqslant 91 e, \\
e \leqslant \frac{91^{2}}{4}2007
$$

pairs of points, which are not connected by an edge. Therefore, the corresponding angles do not exceed $120^{\circ}$.
In conclusion, the minimum value of $n$ is 91.
(Xiong Bin, Gu Hongda, Li Dayuan, Liu Hongkun, Ye Shengyang)