4. (1989 21st Canadian Mathematical Olympiad) Given the numbers $1,2,2^{2}, \cdots, 2^{n-1}$ for any permutation $\sigma=\left(x_{1}\right.$, $\left.x_{2}, \cdots, x_{n}\right)$, define $S_{1}(\sigma)=x_{1}, S_{2}(\sigma)=x_{1}+x_{2}, S_{3}(\sigma)=x_{1}+x_{2}+x_{3}, \cdots, S_{n}(\sigma)=x_{1}+x_{2}+\cdots+x_{n}$, and let $Q(\sigma)=S_{1}(\sigma) \cdot S_{2}(\sigma) \cdot \cdots \cdot S_{n}(\sigma)$, find $\sum \frac{1}{Q(\sigma)}$ (the sum is taken over all permutations).
4. We prove a more general conclusion:

For any $n$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, the sum $\Sigma \frac{1}{Q(\sigma)}=S\left(a_{1}, a_{2}, \cdots, a_{n}\right)=\frac{1}{a_{1} a_{2} \cdots a_{n}}$. We use induction on $n$.
When $n=1$, $S\left(a_{1}\right)=\sum \frac{1}{Q(\sigma)}=\frac{1}{a_{1}}$, so the conclusion holds for $n=1$.
Assume the conclusion holds for $n=k$. We will prove it for $n=k+1$.
In fact, the $(k+1)$ numbers $a_{1}, a_{2}, \cdots, a_{k}, a_{k+1}$ can be divided into $k+1$ classes, each containing $k!$ permutations that end with the same element $a_{i}$. For such permutations, the last factor of $Q(\sigma)$ is $S_{k+1}(\sigma)=a_{1}+a_{2}+\cdots+a_{k}+a_{k+1}$. The first $k$ factors are given by all possible permutations of the other $k$ elements.
By the induction hypothesis, the sum of $\frac{1}{Q(\sigma)}$ for these $k!$ permutations is
$$
\frac{a_{i}}{a_{1} a_{2} \cdots a_{k} a_{k+1} S_{k+1}(\sigma)} \text {. }
$$

Summing over $i$ gives
$$
\begin{aligned}
\sum \frac{1}{Q(\sigma)} & =S\left(a_{1}, a_{2}, \cdots, a_{k+1}\right) \\
& =\frac{1}{a_{1} a_{2} \cdots a_{k} a_{k+1} S_{k+1}(\sigma)} \cdot \sum_{i=1}^{k+1} a_{i} \\
& =\frac{1}{a_{1} a_{2} \cdots a_{k} a_{k+1}} .
\end{aligned}
$$

By the principle of mathematical induction, the general conclusion holds.
Thus, the result for this problem is
$$
\sum \frac{1}{Q(\sigma)}=\frac{1}{1 \cdot 2 \cdot 2^{2} \cdots 2^{n-1}}=2^{-\frac{n(n-1)}{2}} .
$$