24. Let $A=\{1,2,3, \cdots, 4 n+2\}, M=\{2 n+1,4 n+3,6 n+5\}$. For any non-empty subset $B$ of $A$, if the sum of any two numbers in $B$ does not belong to $M$, then $B$ is called an $M$-free set. If $A=A_{1} \cup A_{2}, A_{1} \cap A_{2}=$ $\varnothing$, and $A_{1}, A_{2}$ are both $M$-free sets, then the ordered pair $\left(A_{1}, A_{2}\right)$ is called an $M$-partition of $A$. Try to find the number of all $M-$ partitions of $A$.
24. For $m, p \in A$, if $m+p \in M$, then $m$ and $p$ are said to be related. It is easy to see that the numbers related to 1 are only $2n$ and $4n+2$; the numbers related to $2n$ and $4n+2$ are 1 and $2n+3$; the numbers related to $2n+3$ are $2n$ and $4n+2$. Therefore, $1, 2n+3, 2n, 4n+2$ are divided into two groups $\{1, 2n+3\}, \{2n, 4n+2\}$, forming an unordered pair $(\{1, 2n+3\}, \{2n, 4n+2\})$. Similarly, other groups can be divided: $(\{2, 2n+4\}, \{2n-1, 4n+1\}), (\{3, 2n+5\}, \{2n-2, 4n\}), \cdots, (\{n, 3n+2\}, \{n+1, 3n+3\}), (\{2n+1\}, \{2n+2\})$, a total of $2n+2$ groups, forming $n+1$ unordered pairs (the last pair has only one number in each group). Since any number is related to and only to the numbers in the other group of the same pair, one group in any pair must be in $A_1$ and the other group must be in $A_2$. Conversely, this is also true, and there are no related numbers within $A_1$ and $A_2$ respectively, so the number of $M-$ partitions of $A$ is $2^{n+1}$.