Example 2: Twelve acrobats numbered $1, 2, \cdots, 12$ are divided into two groups, $A$ and $B$, each with six people. Let the actors in group $A$ form a circle, and each actor in group $B$ stand on the shoulders of two adjacent actors in the $A$ circle. If the number of each actor in the $B$ circle is equal to the sum of the numbers of the two actors beneath him, then such a combination is called a "tower". How many structurally different towers can be built?

[Note] The six people in group $A$, $a_{1}, a_{2}, \cdots, a_{6}$, forming a circle in this order in a clockwise or counterclockwise direction can be considered the same structure. A circle is also considered the same structure after rotation. For convenience, draw a circle, fill in the numbers of the bottom actors inside the circle, and fill in the numbers of the top actors outside the circle. Figure 20 is a tower of 14 people.
Let the sum of the elements in groups $A$ and $B$ be denoted as $x$ and $y$ respectively.
Then $y = 2x$.
Therefore, $3x = x + y = 1 + 2 + \cdots + 12 = 78 \Rightarrow x = 26$.
Clearly, $1, 2 \in A, 11, 12 \in B$.
Let $A = \{1, 2, a, b, c, d\} (a < b < c < d)$.
If $d \leq 7$, then
$$
a + b + c + d \leq 4 + 5 + 6 + 7 = 22.
$$

Thus, $a + b + c + d = 23$, and $a \geq 3, 8 \leq d \leq 10$.
(1) $d = 8$.

Then $A = \{1, 2, a, b, c, 8\}, c \leq 7, a + b + c = 15$.
Thus, $(a, b, c) = (3, 5, 7)$ or $(4, 5, 6)$, i.e.,
$A = \{1, 2, 3, 5, 7, 8\}$ or $A = \{1, 2, 4, 5, 6, 8\}$.
(i) If $A = \{1, 2, 3, 5, 7, 8\}$, then
$B = \{4, 6, 9, 10, 11, 12\}$.
Since set $B$ contains $4, 6, 11, 12$, set $A$ must have 1 adjacent to 3, 1 adjacent to 5, 5 adjacent to 7, and 8 adjacent to 3.

In this case, there is only one arrangement, resulting in the tower in Figure 21.
Use 21
(ii) If $A = \{1, 2, 4, 5, 6, 8\}$, then
$$
B = \{3, 7, 9, 10, 11, 12\}.
$$

Similarly, set $A$ must have 1 adjacent to 2, 5 adjacent to 6, and 4 adjacent to 8. In this case, there are two arrangements, resulting in the two towers in Figure 22.
(2) $d = 9$.
Then $A = \{1, 2, a, b, c, 9\}, c \leq 8, a + b + c = 14$.
Thus, $(a, b, c) = (3, 5, 6)$ or $(3, 4, 7)$, i.e.,
$A = \{1, 2, 3, 5, 6, 9\}$ or $A = \{1, 2, 3, 4, 7, 9\}$.
(i) If $A = \{1, 2, 3, 5, 6, 9\}$, then
$B = \{4, 7, 8, 10, 11, 12\}$.
To get $4, 10, 12$ in set $B$, set $A$ must have 1, 3, 9 all adjacent, which is impossible.
(ii) If $A = \{1, 2, 3, 4, 7, 9\}$, then
$B = \{5, 6, 8, 10, 11, 12\}$.
To get $6, 8, 12$ in set $B$, set $A$ must have 2 adjacent to 4, 1 adjacent to 7, and 9 adjacent to 3. Thus, there are two arrangements, resulting in the two towers in Figure 23.
(3) $d = 10$.
Then $A = \{1, 2, a, b, c, 10\}, c \leq 9, a + b + c = 13$.
Thus, $(a, b, c) = (3, 4, 6)$, i.e.,
$$
\begin{array}{l}
A = \{1, 2, 3, 4, 6, 10\}, \\
B = \{5, 7, 8, 9, 11, 12\}.
\end{array}
$$

To get $8, 9, 11, 12$ in set $B$, set $A$ must have 6 adjacent to 2, 6 adjacent to 3, 10 adjacent to 1, and 10 adjacent to 2. Thus, there is only one arrangement, resulting in the tower in Figure 24.
Therefore, there are 6 structurally different towers, i.e.,
$T(6) = 6$.
As the value of $n$ increases, the number of such arrangements $T(n)$ increases rapidly, making the situation more complex.