Example 6 For any non-empty subset $X$ of the set $M=\{1,2, \cdots, 1000\}$, let $\alpha_{X}$ denote the sum of the maximum and minimum numbers in $X$. Find the arithmetic mean of all such $\alpha_{X}$.
(1991, National High School Mathematics Joint Competition)
【Analysis】According to the problem, the required average is
$$
f=\frac{\sum_{\varnothing \neq X \subseteq M} \alpha_{X}}{2^{1000}-1} \text {. }
$$

The key to solving this is to calculate the sum in the numerator
$$
N=\sum_{\varnothing \neq X \subseteq M} \alpha_{X} \text {. }
$$

To compute such an "unordered sum", one must first choose a "sequence". For this, one can consider the contribution of each element to the total sum and then accumulate it, often referred to as the "contribution method"; or consider pairing two subsets so that their "sum" is easy to calculate.

Solution 1 (Contribution Method) Take any $i \in M$, then $i$ contributes to the sum (1) in two ways: $i$ contributes as the maximum number the number of times the subset $\{i\} \cup P(P \subseteq\{1,2, \cdots, i-1\})$ appears, which is $2^{i-1}$; $i$ contributes as the minimum number the number of times the subset $\{i\} \cup P(P \subseteq\{i+1, i+2, \cdots, 1000\})$ appears, which is $2^{1000-i}$.
Thus, the total contribution of $i$ to the numerator is $i\left(2^{i-1}+2^{1000-i}\right)$, and the cumulative "ordered" sum for equation (1) is
$$
N=\sum_{i=1}^{1000} i\left(2^{i-1}+2^{1000-i}\right) \text {. }
$$

Using the method of summing "arithmetic-geometric series" (or Abel's summation formula) we get
$$
N=1001\left(2^{1000}-1\right) \Rightarrow f=1001 .
$$

Solution 2 (Pairing Method) Take any $\varnothing \neq X \subseteq M$, define
$$
X^{\prime}=\{1001-x \mid x \in X\} \text {. }
$$

If $X=X^{\prime}$, then
$$
\begin{array}{l}
\min X=\min X^{\prime}=1001-\max X \\
\Rightarrow \alpha_{X}=\max X+\min X=1001 . \\
\text { If } X \neq X^{\prime} \text {, then } \\
\min X^{\prime}=1001-\max X \text { and } \max X^{\prime}=1001-\min X \\
\Rightarrow \alpha_{X^{\prime}}=1001 \times 2-\alpha_{X} \\
\Rightarrow \alpha_{X}+\alpha_{X^{\prime}}=1001 \times 2 .
\end{array}
$$

If $X \neq X^{\prime}$, then
$$
\begin{array}{l}
\min X^{\prime}=1001-\max X \text { and } \max X^{\prime}=1001-\min X \\
\Rightarrow \alpha_{X^{\prime}}=1001 \times 2-\alpha_{X} \\
\Rightarrow \alpha_{X}+\alpha_{X^{\prime}}=1001 \times 2 .
\end{array}
$$

Let $k$ be the number of non-empty subsets among the $2^{1000}-1$ non-empty subsets that satisfy $X^{\prime}=X$. Then
$$
\begin{array}{l}
N=\sum_{\varnothing \neq X \subseteq M} \alpha_{X} \\
=1001 k+\frac{2^{1000}-1-k}{2} \times 2002 \\
=1001\left(2^{1000}-1\right) . \\
\text { Hence } f=\frac{N}{2^{1000}-1}=1001 .
\end{array}
$$