Method using modular arithmetic
Notice that $\triangle AOB$ is isosceles and therefore $\angle AOB = 105^\circ$.
Each successive chord adds $105^\circ$ to this angle so we want to know when this would be a multiple of $360^\circ$. In other words, we want
$$
105n \equiv 0 \pmod {360}
$$
This implies
$$
7n \equiv 0 \pmod { 24}
$$
Since $\gcd{(7, 24)} = 1$, the minimum $n$ that satisfies this equation is $\boxed{n = 24}$
Method using Complex Numbers
Without loss of generality, assume that the outer circle has a radius of $1$. We can rotate the initial diagram such that the point $A$ corresponds to the point $(1, 0)$ on the coordinate plane.
Every time we draw a chord, the angle increases by $105^\circ$, so essentially we are rotating the complex number by $105^\circ$ each time.
Let $z = e^{i \cdot 0}$. The next chord point (in our case $B$) will occour at $z = e^{i \cdot \frac{7\pi}{12}}$ and $C$ will occour at $z = e^{i \cdot \frac{7\pi}{6}}$ and so on.
We want to know, when $z = e^{i \cdot \frac{7\pi}{12} \cdot n}$ will return to the point $A$ at $(1, 0)$. Clearly, this will happen when:
$$
\frac{7\pi}{12}\cdot n = 2\pi k
$$
In other words, for some integer $n$ we want:
$$
7n = 24k
$$
The above simply corresponds to the modular equation:
$$
7n \equiv 0 \pmod {24}
$$
Since $\gcd(7,24)=1$, the minimum n that satisfies this equation is $\boxed{n=24}$.
Verification
If we use complex numbers, we actually know the coordinates of each point. For example,
$$
e^{i \cdot \frac{7\pi}{12}} \implies (\cos(105^\circ), \ \sin(105^\circ))
$$
Using this we can be sure that the only solution occurs when $n = 24$.