15 Method 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 $A=\left\{a_{1}, a_{2}, \cdots, a_{6}\right\}$, 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
$$
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=27,
$$

so $A_{1}=\left\{a_{1}, a_{2}, a_{3}, a_{1}, a_{5}\right\}$ 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 2 even numbers, let $p$ represent the sum of the 2 even numbers $x_{1}, x_{2}$ in $A_{1}$, and $q$ represent the sum of the 3 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 the following 24 scenarios for $A_{1}$.
( $1^{\circ}$ ) When $p=6, q=21$, then $\left(x_{1}, x_{2}\right)=(2,4),\left(y_{1}, y_{2}, y_{3}\right)$ can be $(1,9,11),(3,7,11),(5,7,9)$, thus there are 3 scenarios for $A_{1}$;
( $\left.2^{\circ}\right)$ When $p=8, q=19$, then $\left(x_{1}, x_{2}\right)=(2,6),\left(y_{1}, y_{2}, y_{3}\right)$ can be $(1,7,11),(3,5,11),(3,7,9)$, thus there are 3 scenarios for $A_{1}$;
(3 $\left.3^{\circ}\right)$ When $p=10, q=17$, then $\left(x_{1}, x_{2}\right)$ can be $(2,8),(4,6),\left(y_{1}, y_{2}, y_{3}\right)$ can be $(1,5,11),(1,7,9),(3,5,9)$, thus there are 6 scenarios for $A_{1}$;
(4 $\left.4^{\circ}\right)$ When $p=12, q=15$, then $\left(x_{1}, x_{2}\right)$ can be $(2,10),(4,8),\left(y_{1}, y_{2}, y_{3}\right)$ can be $(1,3,11),(1,5,9),(3,5,7)$, thus there are 6 scenarios for $A_{1}$;
( $\left.5^{\circ}\right)$ When $p=14, q=13$, then $\left(x_{1}, x_{2}\right)$ can be $(4,10),(6,8),\left(y_{1}, y_{2}, y_{3}\right)$ can be $(1,3,9),(1,5,7)$, thus there are 4 scenarios for $A_{1}$;
( $6^{\circ}$ ) When $p=16, q=11$, then $\left(x_{1}, x_{2}\right)=(6,10),\left(y_{1}, y_{2}, y_{3}\right)=(1, 3,7)$, thus there is 1 scenario for $A_{1}$;
$\left(7^{\circ}\right)$ When $p=18, q=9$, then $\left(x_{1}, x_{2}\right)=(8,10),\left(y_{1}, y_{2}, y_{3}\right)=(1, 3,5)$, thus there is 1 scenario for $A_{1}$.
(3) If $A_{1}$ contains 1 odd number and 4 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 find that there are $1+24+4=29$ (types) of sets $A$, i.e., $M$ has 29 "equal sum partitions".
Method 2: Element exchange method, clearly $\sum_{i=1}^{6} a_{i}=\sum_{i=1}^{6} b_{i}=39$, always assume $12 \in A$.
First, consider an extreme case of partition:
$$
\begin{array}{c}
A_{0}=\{1,2,3,10,11,12\}, \\
B_{0}=\{4,5,6,7,8,9\},
\end{array}
$$

Obviously, if one set of numbers in the arrays $\{1,2,3\}$ and $\{10,11,12\}$ is all in $A$, then the other set must also be all in $A$; the following considers 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 in $A_{0}$ and $B_{0}$:
$\left(1^{\circ}\right)(10,1) \leftrightarrow(5,6),(4,7) ;(10,2) \leftrightarrow(5,7),(4,8) ;(10,3) \leftrightarrow$ $(6,7),(5,8),(4,9) ;(10,2,3) \leftrightarrow(4,5,6)$, which results in 8 exchanges;
(2 $)(11,1) \leftrightarrow(5,7),(4,8) ;(11,2) \leftrightarrow(6,7),(5,8),(4,9)$; $(11,3) \leftrightarrow(6,8),(5,9) ;(11,1,3) \leftrightarrow(4,5,6) ;(11,2,3) \leftrightarrow(4,5, 7)$, which results in 9 exchanges;
$\left(3^{\circ}\right)(10,11,1) \leftrightarrow(6,7,9),(5,8,9) ;(10,11,2) \leftrightarrow(6,8,9) ;$ $(10,11,3) \leftrightarrow(7,8,9) ;(10,11,1,2) \leftrightarrow(4,5,7,8),(4,5,6,9)$; $(10,11,1,3) \leftrightarrow(4,6,7,8),(4,5,7,9) ;(10,11,2,3) \leftrightarrow(5,6,7, 8)$, (4, 6, 7, 9), (4, 5, 8, 9), which results in 11 exchanges.

Each exchange results in a new partition, therefore, this problem has a total of $1+8+9+11=29$ (types) of "equal sum partitions".
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 $M$ (considering $A \cup B$ and $B \cup A$ as the same partition).
Determine the number of “equal-sum partitions” for the set $M=\{1,2, \cdots, 12\}$.