The answer is 1001.
Let $S=\{1,2, \cdots, n\}, n \in N$. Let $m_{A}, M_{A}$ represent the minimum and maximum elements of a non-empty subset $A$ of $S$, respectively. Then the arithmetic mean of all $\alpha_{A}$ is
$$\frac{1}{2^{n}-1} \sum_{\substack{A \subset S \\ A \neq \varnothing}}\left(m_{A}+M_{A}\right).$$

Since for $1 \leqslant k \leqslant n$, there are $2^{n-k}$ subsets $A$ satisfying $m_{A}=k$, we have
$$\begin{array}{l} \sum_{\substack{A \subset S \\ A \neq \infty}} m_{A}=1 \cdot 2^{n-1}+2 \cdot 2^{n-2}+\cdots+ \\ k \cdot 2^{n-1}+\cdots+n \cdot 2^{0}. \end{array}$$

And for $1 \leqslant k \leqslant n$, there are $2^{k-1}$ subsets $A$ such that $M_{A}=k$, hence
$$\begin{array}{l} \sum_{\substack{A}} M_{A}=n \cdot 2^{n-1}+(n-1) 2^{n-2}+\cdots \\ \quad+(n-k+1) 2^{n-k}+\cdots+1 \cdot 2^{0}.(3) \end{array}$$
Adding (2) and (3) gives
$$\begin{array}{l} \sum_{A \subseteq S}\left(m_{A}+M_{A}\right)=(n+1)\left(2^{n-1}+2^{n-2}\right. \\ A \neq \varnothing \\ +\cdots+2+1)=(n+1)\left(2^{n}-1\right). \end{array}$$

Combining (1), we know that the arithmetic mean of all such $\alpha_{A}$ is $n+1$.
From (2), we get
$$\begin{array}{l} 2 \sum_{A \subset S} m_{A}=2^{n}+2 \cdot 2^{n-1}+\cdots \\ A \neq \varnothing \\ \quad+(t+1) 2^{n-I}+\cdots+n \cdot 2. \end{array}$$
Subtracting (2) from (A), we get
$$\begin{aligned} \sum_{\substack{A \subset S}}^{n_{A} \neq \varnothing} & =2^{n}+2^{n-1}+\cdots+2^{n-x}+2-n \\ & =2^{u+1}-n-2. \end{aligned}$$

Similarly, we have
$$\sum_{\substack{A \subset S \\ A \neq \varnothing}} M_{\triangle}=(n-1) 2^{\mathbb{n}}+1$$

Finally, let $1 \leqslant i<i \leqslant n$, then the number of subsets $A \subset S$ with $i$ and $1$ as the minimum and maximum elements, respectively, is $2^{i-1-1}$, and in this case, $\alpha_{A}=i+j$. Noting that $\alpha_{\{1\}}=2 i, 1 \leqslant i \leqslant n$. Therefore, we have the identity
$$\begin{array}{l} \mathbf{2} \sum_{\mathbf{1}=1}^{\mathbf{n}} i+\sum_{1 \leqslant \mathrm{i}<\mathbf{j} \leqslant \mathbf{n}}(i+j) 2^{i-1-1} \\ =(n+1) \cdot\left(2^{\mathbf{n}}-1\right), \end{array}$$

or $\sum_{1 \leqslant \mathrm{i}<\mathrm{j}=\mathrm{n}}(i+1) 2^{\mathrm{j}-1}=2(n+1)\left(2^{\mathrm{n}}-n-1\right)$ (8)