15. 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 $M$ ($A \cup B$ and $B \cup A$ are considered the same partition). Determine the number of equal-sum partitions of the set $M=\{1,2, \cdots, 12\}$.
15. Solution 1: Without loss of generality, let $12 \in A$. Since when set $A$ is determined, set $B$ is uniquely determined, we only need to consider the number of sets $A$.
Let $A=\left\{a_{1}, a_{2}, \cdots, a_{6}\right\}, a_{6}$ be the largest number.
By $1+2+\cdots+12=78$, we know
$a_{1}+a_{2}+\cdots+a_{6}=39, a_{6}=12$.
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 five odd numbers, since the sum of the six odd numbers in $M$ is 36, and $27=36-9$, we have
$A_{1}=\{1,3,5,7,11\}$
At this point, we get the unique $A=\{1,3,5,7,11,12\}$.
(2) If $A$ contains three odd numbers and two even numbers, let $p$ 

represent the sum of the two even numbers $x_{1}, x_{2}$ in $A_{1}$, and $q$ 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, $q \leqslant 21, p \leqslant 18$.
We get 24 cases for $A_{4}$.
(1) When $p=6, q=21$,
$\left(x_{1}, x_{2}\right)=(2,4)$,
$\left(y_{1}, y_{2}, y_{3}\right)=(1,9,11),(3,7,11),(5,7,9)$,

we get 3 cases for $A_{1}$;
(2) When $p=8, q=19$,
$\left(x_{1}, x_{2}\right)=(2,6)$,
$\left(y_{1}, y_{2}, y_{3}\right)=(1,7,11),(3,5,11),(3,7,9)$,

we get 3 cases for $A_{1}$;
(3) When $p=10, q=17$,
$\left(x_{1}, x_{2}\right)=(2,8),(4,6)$,
$\left(y_{1}, y_{2}, y_{3}\right)=(1,5,11),(1,7,9),(3,5,9)$,

we get 6 cases for $A_{1}$;
(4) When $p=12, q=15$,
$\left(x_{1}, x_{2}\right)=(2,10),(4,8)$,
$\left(y_{1}, y_{2}, y_{3}\right)=(1,3,11),(1,5,9),(3,5,7)$,

we get 6 cases for $A_{1}$;
(5) When $p=14, q=13$,
$\left(x_{1}, x_{2}\right)=(4,10),(6,8)$,
$\left(y_{1}, y_{2}, y_{3}\right)=(1,3,9),(1,5,7)$,

we get 4 cases for $A_{1}$;
(6) When $p=16, q=11$,
$\left(x_{1}, x_{2}\right)=(6,10),\left(y_{1}, y_{2}, y_{3}\right)=(1,3,7)$,

we get 1 case for $A_{1}$;
(7) When $p=18, q=9$,
$\left(x_{1}, x_{2}\right)=(8,10),\left(y_{1}, y_{2}, y_{3}\right)=(1,3,5)$,

we get 1 case for $A_{1}$.
(3) If $A_{1}^{\prime}$ 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 cases for $A_{1}$
$(7,2,4,6,8),(5,2,4,6,10)$,
$(3,2,4,8,10),(1,2,6,8,10)$.
In summary, there are $1+24+4=29$ cases for set $A$,

i.e., $M$ has 29 equal-sum partitions.
Solution 2: Element exchange method.
Clearly, $\sum_{i=1}^{6} a_{i}=\sum_{i=1}^{6} \cdot b_{i}$, always assume $12 \in A$.
(1) First, note one extreme partition:
$A_{0}=\{1,2,3,10,11,12\}$,
$B_{0}=\{4,5,6,7,8,9\}$.
Clearly, if the set $\{1,2,3\}$ is entirely in $A$, then the set $\{10,11,12\}$ must also be entirely in $A$.

Next, consider the case where at least one of the numbers 10, 11 is not in $A$. For this, consider element exchanges of the same number and equal sum between $A_{0}$ and $B_{0}$.
(2) $(10,1) \leftrightarrow(5,6),(4,7)$;
$(10,2) \leftrightarrow(5,7),(4,8)$;
$(10,3) \leftrightarrow(6,7),(5,8),(4,9)$;
$(10,2,3) \leftrightarrow(4,5,6)$.
We get 8 exchanges.
(3) $(11,1) \leftrightarrow(5,7),(4,8)$;
$(11,2) \leftrightarrow(6,7),(5,8),(4,9)$;
$(11,3) \leftrightarrow(6,8),(5,9)$;
$(11,1,3) \leftrightarrow(4,5,6)$;
$(11,2,3) \leftrightarrow(4,5,7)$.
We get 9 exchanges.
(4) $(10,11,1) \leftrightarrow(6,7,9),(5,8,9)$;
$(10,11,2) \leftrightarrow(6,8,9)$;
$(10,11,3) \leftrightarrow(7,8,9)$;
$(10,11,1,2) \leftrightarrow(4,5,7,8),(4,5,6,9)$;
$(10,11,1,3) \leftrightarrow(4,6,7,8),(4,5,7,9)$;
$(10,11,2,3) \leftrightarrow(5,6,7,8),(4,6,7,9)$,
$(4,5,8,9)$.
We get 11 exchanges.
Each exchange results in a new partition.
Therefore, we have a total of $1+8+9+11=29$ equal-sum partitions.