$2 \cdot 63$ Let the set $M=\{1,2, \cdots, 1000\}$. For any non-empty subset $X$ of $M$, let $\alpha_{X}$ denote the sum of the largest and smallest numbers in $X$. Find the arithmetic mean of all such $\alpha_{Y}$.
[Solution 1] Pair the non-empty subsets of set $M$: For each non-empty set $X \subset M$, let
$$
X^{\prime}=\{1001-x \mid x \in X\},
$$

then when $M \supset X_{1} \neq X$, $X^{\prime} \neq X^{\prime}$. Thus, all non-empty subsets can be divided into two categories: (1) $\left\{X \mid X \subset M, X^{\prime} \neq X\right\}$; (2) $\left\{X \mid X \subset M, X^{\prime}=X\right\}$.

Clearly, for the $X$ in the second category, $\alpha_{X}=1001$. And for a pair $X$ and $X^{\prime}$ in the first category, $\alpha_{X}+\alpha_{X}^{\prime}=2002$. Therefore, the arithmetic mean of all $\alpha_{X}$ is 1001.
[Solution 2] First, for each $i \in M$, we calculate the number of sets for which it is the maximum number and the number of sets for which it is the minimum number. It is easy to see that these two numbers are $2^{i-1}$ and $2^{1000-i}$, respectively. Therefore, the sum of all $\alpha_{X}$ is
$$
S=\sum_{i=1}^{1000}\left(2^{i-1}+2^{1000-i}\right) i .
$$

By adding the sum $S$ to its reverse sum, we get
$$
\begin{aligned}
S & =\frac{1}{2}\left\{\sum_{i=1}^{1000}\left(2^{i-1}+2^{1000-i}\right) i+\sum_{i=1}^{1000}\left(2^{1000-i}+2^{i-1}\right)(1001-i)\right\} \\
& =1001 \sum_{i=1}^{1000}\left(2^{i-1}+2^{1000-i}\right) .
\end{aligned}
$$

Since the sum on the right side of the above equation is exactly the total number of all non-empty subsets, it follows that the arithmetic mean of all $\alpha_{X}$ is 1001.