Four, (50 points) For the $2n$-element set $M=\{1,2, \cdots, 2n\}$, if the $n$-element sets $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}, B=$ $\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ satisfy: $A \cup B=M, A \cap B=\varnothing$, and $\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} b_{k}$, then $A \cup B$ is called an "equal-sum partition" of set $M$ (considering $A \cup B$ and $B \cup A$ as the same partition). Determine how many "equal-sum partitions" the set $M=\{1,2, \cdots, 12\}$ has.
Solution 1: Let's assume $12 \in A$. Since when set $A$ is determined, set $B$ is uniquely determined, we only need to consider the number of sets $A$. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{6}\right\}$, with $a_{6}$ being the largest number. Given $1+2+\cdots+12=78$, then
$a_{1}+a_{2}+\cdots+a_{6}=39, a_{6}=12$, thus $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=27$,
hence $A_{1}=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$ contains an odd number of odd numbers.
(1) If $A_{1}$ contains 5 odd numbers, since the sum of the six odd numbers in $M$ is 36, and $27=36-9$, then $A_{1}=\{1,3,5,7,11\}$, which gives the unique $A=\{1,3,5,7,11,12\}$;
(2) If $A_{1}$ contains 3 odd numbers and two even numbers; let $p$ represent the sum of the two even numbers $x_{1}, x_{2}$ in $A_{1}$; $q$ represent the sum of the three odd numbers $y_{1}, y_{2}, y_{3}$ in $A_{1}$, then $p \geqslant 6, q \geqslant 9$, thus $q \leqslant 21, p \leqslant 18$. This results in 24 scenarios for $A_{1}$.

Among them, $\left(1^{0}\right)$ when $p=6, q=21$, then $\left(x_{1}, x_{2}\right)=(2,4),\left(y_{1}, y_{2}, y_{3}\right)=(1,9,11),(3,7,11),(5,7,9)$; this results in 3 scenarios for $A_{1}$;
$\left(2^{\circ}\right)$ when $p=8, q=19$, then $\left(x_{1}, x_{2}\right)=(2,6),\left(y_{1}, y_{2}, y_{3}\right)=(1,7,11),(3,5,11),(3,7,9)$; this results in 3 scenarios for $A_{1}$;
(3) when $p=10, q=17$, then $\left(x_{1}, x_{2}\right)=(2,8),(4,6),\left(y_{1}, y_{2}, y_{3}\right)=(1,5,11),(1,7,9),(3,5,9)$, this results in 6 scenarios for $A_{1}$;
$\left(4^{\circ}\right)$ when $p=12, q=15$, then $\left(x_{1}, x_{2}\right)=(2,10),(4,8),\left(y_{1}, y_{2}, y_{3}\right)=(1,3,11),(1,5,9),(3,5,7)$, this results in 6 scenarios for $A_{1}$;
( $\left.5^{\circ}\right)$ when $p=14, q=13$, then $\left(x_{1}, x_{2}\right)=(4,10),(6,8),\left(y_{1}, y_{2}, y_{3}\right)=(1,3,9),(1,5,7)$, this results in 4 scenarios for $A_{1}$;
$\left(6^{\circ}\right)$ when $p=16, q=11$, then $\left(x_{1}, x_{2}\right)=(6,10),\left(y_{1}, y_{2}, y_{3}\right)=(1,3,7)$; this results in 1 scenario for $A_{1}$;
$\left(7^{\circ}\right)$ when $p=18, q=9$, then $\left(x_{1}, x_{2}\right)=(8,10),\left(y_{1}, y_{2}, y_{3}\right)=(1,3,5)$; this results in 1 scenario for $A_{1}$.
(3) If $A_{1}$ contains one odd number and four even numbers, since the sum of the five even numbers in $M$ except 12 is $2+4+6+8+10=30$, removing one even number and adding one odd number to make the sum of the five numbers in $A_{1}$ equal to 27, we get 4 scenarios for $A_{1}$: $(7,2,4,6,8)$, $(5,2,4,6,10),(3,2,4,8,10),(1,2,6,8,10)$.
Combining the above three steps, we know that set $A$ has $1+24+4=29$ scenarios, i.e., $M$ has 29 "equal-sum partitions".
Solution 2: Element exchange method, clearly $\sum_{i=1}^{6} a_{i}=\sum_{i=1}^{6} b_{i}=39$, always assume $12 \in A$;
$\left(1^{\circ}\right)$ First, note the extreme case of a partition: $A_{0}=\{1,2,3,10,11,12\}, B_{0}=\{4,5,6,7,8,9\}$, obviously the array $\{1,2,3\}$ and $\{10,11,12\}$, if one set of numbers is all in $A$, then the other set of numbers must all be in $A$;

Next, consider the case where at least one of the numbers 10, 11 is not in $A$. For this, consider the element exchanges of the same number and equal sum in $A_{0}, B_{0}$:
$$
\left(2^{\circ}\right)(10,1) \leftrightarrow(5,6),(4,7) ;(10,2) \leftrightarrow(5,7),(4,8) ;(10,3) \leftrightarrow(6,7),(5,8),(4,9) ;(10,2,3) \leftrightarrow
$$
$(4,5,6)$; this results in 8 exchanges;
$$
\left(3^{\circ}\right)(11,1) \leftrightarrow(5,7),(4,8) ;(11,2) \leftrightarrow(6,7),(5,8),(4,9) ;(11,3) \leftrightarrow(6,8),(5,9) ;(11,1,3) \leftrightarrow
$$
$(4,5,6) ;(11,2,3) \leftrightarrow(4,5,7)$; this results in 9 exchanges;
$$
\begin{array}{l}
\left(4^{0}\right)(10,11,1) \leftrightarrow(6,7,9),(5,8,9) ;(10,11,2) \leftrightarrow(6,8,9) ;(10,11,3) \leftrightarrow(7,8,9) ; \\
(10,11,1,2) \leftrightarrow(4,5,7,8),(4,5,6,9) ;(10,11,1,3) \leftrightarrow(4,6,7,8),(4,5,7,9) ;
\end{array}
$$
$(10,11,2,3) \leftrightarrow(5,6,7,8),(4,6,7,9),(4,5,8,9)$; this results in 11 exchanges. Each exchange results in a new partition, therefore, this problem has $1+8+9+11=29$ equal-sum partitions.