6 Let the arithmetic mean of all elements in the set $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be denoted as $P(A)\left(P(A)=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)$. If $B$ is a non-empty subset of $A$ and $P(B)=P(A)$, then $B$ is called a “balanced subset” of $A$. Try to find the number of all “balanced subsets” of the set $M=\{1,2,3,4,5,6,7,8,9\}$. (Provided by Tao Pingsheng)
Given $P(M)=5$, let $M^{\prime}=\{x-5 \mid x \in M\}=\{-4,-3,-2,-1,0,1,2,3,4\}$, then $P\left(M^{\prime}\right)=0$. According to this translation relationship, the balanced subsets of $M$ and $M^{\prime}$ can be one-to-one corresponded. Let $f(k)$ denote the number of $k$-element balanced subsets of $M^{\prime}$. Clearly, $f(9)=1, f(1)=1$ (the 9-element balanced subset of $M^{\prime}$ is only $M^{\prime}$, and the 1-element balanced subset is only $\{0\}$).

The two-element balanced subsets of $M^{\prime}$ are four, namely $B_{i}=\{-i, i\}, i=1,2,3,4$, so $f(2)=4$.

The three-element balanced subsets of $M^{\{\prime\}}$ fall into two cases:
(1) Those containing the element 0 are $B_{i} \cup\{0\}=\{-i, 0, i\}, i=1,2,3,4$, totaling four;
(2) Those not containing the element 0, due to the equations $3=1+2,4=1+3$ which can be expressed as $-3+1+2=0,3-1-2=0$ and $-4+1+3=0,4-1-3=0$, yield 4 balanced subsets $\{-3,1,2\},\{3,-1,-2\},\{-4,1,3\}$, $\{4,-1,-3\}$, so $f(3)=4+4=8$.

The four-element balanced subsets of $M^{\prime}$ fall into three cases:
(1) The union of every two two-element balanced subsets: $B_{i} \cup B_{j}, 1 \leqslant i<j \leqslant 4$, totaling 6 sets;
(2) The union of three-element balanced subsets not containing the element 0 with $\{0\}$, totaling 4 sets;
(3) All other cases, due to the equation $1+4=2+3$ which can be expressed as $-1-4+2+3=0$ and $1+4-2-3=0$, yield 2 balanced subsets $\{-1,-4,2,3\}$ and $\{1,4,-2,-3\}$, so $f(4)=6+4+2=12$.

Also note that, except for $M^{\prime}$ itself, if $B^{\prime}$ is a balanced subset of $M^{\prime}$, then its complement $C_{M^{\prime}} B^{\prime}$ is also a balanced subset of $M^{\prime}$, and they correspond one-to-one. Therefore, $f(9-k)=f(k), k=1,2,3,4$.
Thus, the number of balanced subsets of $M^{\prime}$ is
$$
\begin{aligned}
\sum_{k=1}^{9} f(k) & =f(9)+2 \sum_{k=1}^{4} f(k) \\
& =1+2(1+4+8+12)=51 .
\end{aligned}
$$

That is, $M$ has 51 balanced subsets.