5. Find the smallest positive integer $n$ such that: for any $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ taken on the circumference of circle $O$, among the $\mathrm{C}_{n}^{2}$ angles $\angle A_{i} O A_{j}(1 \leqslant i<j \leqslant n)$, at least 2007 angles are not greater than $120^{\circ}$.
5. Solution: First, when $n=90$, as shown in the figure, let $AB$ be the diameter of $\odot O$, and take 45 points near points $A$ and $B$ respectively. At this time, there are only $2 \mathrm{C}_{45}^{2}=1980$ angles not exceeding $120^{\circ}$, so it does not meet the requirement;

When $n=91$, we will prove that there are at least 2007 angles not exceeding $120^{\circ}$.

For 91 points on the circumference, if $\angle A_{i} O A_{j} > 120^{\circ}$, then connect $A_{i} A_{j}$. This way, we get a graph with $e$ edges.

Since $\angle A_{i} O A_{j} > 120^{\circ}$, $\angle A_{i} O A_{k} < 120^{\circ}$, and $\angle A_{j} O A_{k} < 120^{\circ}$, the graph has no triangles. Therefore, the number of angles not exceeding $120^{\circ}$ is $\mathrm{C}_{91}^{2} - e = 4095 - e$. To ensure that there are at least 2007 angles not exceeding $120^{\circ}$, we need $4095 - e \geq 2007$, which implies $e \leq 2088$.

If $e \geq 1$, without loss of generality, assume $A_{1}$ and $A_{2}$ are connected by an edge. Since the graph has no triangles, for any point $A_{i} (3 \leq i \leq 91)$, it can be connected to at most one of $A_{1}$ or $A_{2}$. Therefore, $d(A_{1}) + d(A_{2}) \leq 89 + 2 = 91$, where $d(A)$ represents the number of edges from $A$.

Since $d(A_{1}) + d(A_{2}) + \cdots + d(A_{91}) = 2e$, and for each edge in the graph $G$ with endpoints $A_{i}$ and $A_{j}$, we have $d(A_{i}) + d(A_{j}) \leq 91$, summing this up for all edges gives:
$$
(d(A_{1}))^2 + (d(A_{2}))^2 + \cdots + (d(A_{91}))^2 \leq 91e,
$$
By the Cauchy-Schwarz inequality:
$$
91 \left[(d(A_{1}))^2 + (d(A_{2}))^2 + \cdots + (d(A_{91}))^2\right] \geq \left[d(A_{1}) + d(A_{2}) + \cdots + d(A_{91})\right]^2 = 4e^2,
$$
Thus, $\frac{4e^2}{91} \leq (d(A_{1}))^2 + (d(A_{2}))^2 + \cdots + (d(A_{91}))^2 = 91e$, which implies $e \leq \frac{91^2}{4}$. Therefore, there are at least $4095 - 2088 = 2007$ pairs of points that are not connected by an edge, and the corresponding angles are not exceeding $120^{\circ}$.
In conclusion, the minimum value of $n$ is 91.