Example 2 As shown in Figure 3, two equilateral triangles overlap to form a six-pointed star. Now, the positive integers 1, 2, $\cdots, 12$ are to be filled in the 12 nodes of the figure, such that the sum of the four numbers on each straight line is equal.
(1) Try to find the minimum value of the sum of the numbers at the six vertices $a_{1}, a_{2}, \cdots, a_{6}$ of the six-pointed star;
(2) Prove: The number of different filling schemes that meet the conditions is even (if filling scheme $K$ can coincide with filling scheme $T$ after rotation or reflection, they are considered the same scheme).
(1) Solution: For any arrangement that satisfies the conditions, the number filled at point $a_i$ $(i=1,2, \cdots, 12)$ is still denoted as $a_i$. If the sum of the four numbers on each line is $s$, then
$6 s=2(1+2+\cdots+12) \Rightarrow s=26$.
Therefore, in $\triangle a_{1} a_{3} a_{5}$ and $\triangle a_{2} a_{4} a_{6}$, the sums of the numbers on the three sides are respectively
$$
\begin{array}{l}
\left(a_{1}+a_{8}+a_{9}+a_{3}\right)+ \\
\left(a_{3}+a_{10}+a_{11}+a_{5}\right)+ \\
\left(a_{5}+a_{12}+a_{7}+a_{1}\right) \\
=3 s, \\
\left(a_{2}+a_{9}+a_{10}+a_{4}\right)+ \\
\left(a_{4}+a_{11}+a_{12}+a_{6}\right)+ \\
\left(a_{6}+a_{7}+a_{8}+a_{2}\right) \\
=3 s .
\end{array}
$$

Subtracting the two equations gives $a_{1}+a_{3}+a_{5}=a_{2}+a_{4}+a_{6}$.
Thus, the sum of the numbers at the vertices of the two triangles is equal.
If we denote $m=\left(a_{1}+a_{3}+a_{5}\right)+\left(a_{2}+a_{4}+a_{6}\right)$, then $m$ is an even number.
Because in $1,2, \cdots, 12$, the sum of the smallest six numbers is
$1+2+\cdots+6=21$ (an odd number),
so, $m \geqslant 22$.
If $m=22$, since in $1,2, \cdots, 12$, the only six numbers that sum to 22 are $1,2,3,4,5,7$, and dividing them into two groups with equal sums, there is only one way: $\{1,3,7\}$ and $\{2,4,5\}$.

At this point, if we directly attempt to construct or exclude, it will be too many cases to classify and exhaust, often leading to no good end.

Therefore, we reverse the problem (i.e., "reverse engineering") and use the "clustering" method.

Now, ignoring the specific numbers on the outer circle, and changing $\{1,3,7\}$ and $\{2,4,5\}$ to $\{$ odd, odd, odd $\}$ and $\{$ even, even, odd $\}$, as shown in Figure 4. Thus, under rotation, the way to fill the numbers at the six vertices is unique.

Next, consider the way to fill the remaining six numbers, using the notation $a \sim b$ to indicate that integers $a$ and $b$ have the same parity.
Since the sum of the four numbers on each line $s$ is even, then
$a_{9} \sim a_{8} \sim a_{7} \sim a_{12}, a_{10} \sim a_{11}$.
Among the six numbers in the inner circle $6,8,9,10,11,12$, there are exactly four even numbers and two odd numbers, so $\left\{a_{10}, a_{11}\right\}=\{9,11\}$.

Since $a_{10}$ and $a_{11}$ are on one side of the triangle with odd vertices, the sum of the numbers at the two vertices of that side should be $s-(9+11)=6$. However, there are no two numbers in $\{1,3,7\}$ that sum to 6, which is a contradiction. Therefore, $m \neq 22$, and thus, the even number $m \geqslant 24$.
When $m=24$, as shown in Figure 5, a valid filling method can be given.
Therefore, the minimum value sought is 24.
(2) Proof: For any filling scheme $K$ that satisfies the conditions, let the numbers filled at each point be $a_{1}, a_{2}, \cdots, a_{12}$.
Let $b_{i}=13-a_{i}(i=1,2, \cdots, 12)$.
Then when $a_{i} \in\{1,2, \cdots, 12\}$, $b_{i} \in\{1,2, \cdots, 12\}$.
In the filling scheme $K$, replace $a_{1}, a_{2}, \cdots, a_{12}$ with $b_{1}, b_{2}, \cdots, b_{12}$, to get the filling scheme $T$. Then $T$ is also a filling scheme that satisfies the conditions.
Obviously, $\sum_{i=1}^{6} a_{i}+\sum_{i=1}^{6} b_{i}=\sum_{i=1}^{6}\left(a_{i}+b_{i}\right)=78$.
And $\sum_{i=1}^{6} a_{i}$ and $\sum_{i=1}^{6} b_{i}$ are both even numbers, so
$\sum_{i=1}^{6} a_{i} \neq 39, \sum_{i=1}^{6} b_{i} \neq 39$.
Let the set of all filling schemes that satisfy the conditions be $M$, and divide $M$ into two disjoint subsets: $M=M_{1} \cup M_{2}$, where,
$M_{1}$ contains each filling scheme $K_{1}=\left(a_{1}, a_{2}, \cdots, a_{12}\right)$, the sum of the numbers at the six vertex points $\sum_{i=1}^{6} a_{i}39$.
Make the correspondence between the filling scheme $K_{1}=\left(a_{1}, a_{2}, \cdots, a_{12}\right)$ in set $M_{1}$ and the filling scheme $K_{2}=\left(b_{1}, b_{2}, \cdots, b_{12}\right)$ in set $M_{2}$, such that
$$
b_{i}=13-a_{i}(i=1,2, \cdots, 12) \text {, }
$$

This correspondence is one-to-one.
Therefore, $\left|M_{1}\right|=\left|M_{2}\right|, M_{1} \cap M_{2}=\varnothing$.
And $|M|=\left|M_{1} \cup M_{2}\right|=2\left|M_{1}\right|$, i.e., the number of elements in set $M$ is even, which means there are an even number of different filling schemes that satisfy the conditions.