There are $4 n\left(n \in \mathbf{Z}_{+}\right)$ circles with equal radii on a plane, where any two circles do not intersect with the other three circles. Let the number of intersection points of these circles be $f(n)$. Find the minimum value of $f(n)$.
Let $M$ be the set of these $4 n$ circles $C_{1}, C_{2}, \cdots, C_{4 n}$, and $N$ be the set of the intersection points $P_{1}, P_{2}, \cdots, P_{f(n)}$ of these $4 n$ circles.
Let $C \in M$ and $P \in N$.
If point $P$ is not on circle $C$, define $F(C, P)=0$;
If point $P$ is on circle $C$, define $F(C, P)=\frac{1}{m}$, where $m$ is the number of circles passing through point $P$.
Then for any point $P$ in set $N$, we have
$$
\sum_{i=1}^{4 n} F\left(C_{i}, P\right)=1 \text {. }
$$

For any circle $C$ in set $M$, select $P \in N$ and $P \in C$ such that $F(C, P)=\frac{1}{m}$ is the smallest.

Let $C_{1}^{P}, C_{2}^{P}, \cdots, C_{m-1}^{P}$ be the other $m-1$ circles passing through point $P$, where the other intersection point of circle $C_{i}^{P} (i=1,2, \cdots, m-1)$ and circle $C$ is $Q_{i}$. Since the radii of these circles are equal, the $m-1$ points $Q_{i}$ are distinct.
$$
\begin{array}{l}
\text { Then } \sum_{j=1}^{((n)} F\left(C, P_{j}\right) \geqslant F(C, P)+\sum_{i=1}^{m-1} F\left(C, Q_{i}\right) \\
\geqslant \frac{1}{m}+(m-1) \frac{1}{m}=1 . \\
\text { Hence } f(n)=\sum_{j=1}^{(n n)} \sum_{i=1}^{4 n} F\left(C_{i}, P_{j}\right) \\
=\sum_{i=1}^{4 n} \sum_{j=1}^{4(n)} F\left(C_{i}, P_{j}\right) \geqslant \sum_{i=1}^{4 n} 1=4 n .
\end{array}
$$

It is only necessary to show that $f(n)=4 n$ can be achieved.
As shown in Figure 2, every four
circles have 4 intersection points
(essentially, all circles
have a radius of $r$, and the side length of
the equilateral $\triangle P Q R$ is
$\sqrt{3} r$, these four circles are
the circumcircles of $\triangle P Q R$, $\triangle S P Q$,
$\triangle S Q R$, and $\triangle S R P$).
Therefore, $f_{\text {min }}(n)=4 n$.