$2 \cdot 31$ Let $\sigma(S)$ denote the sum of all elements in a non-empty integer set $S$. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{11}\right\}$ be a set of positive integers, and $a_{1}<a_{2}<\cdots<a_{11}$. If for each positive integer $n \leqslant 1500$, there exists a subset $S$ of $A$ such that $\sigma(S)=n$. Find the minimum value of $a_{10}$ that satisfies the above requirement.
Let $S_{k}=a_{1}+a_{2}+\cdots+a_{k}(1 \leqslant k \leqslant 11)$. By the problem statement, there exists an index $m$ such that $S_{m-1}S_{k-1}+1$, and $a_{1}+\cdots+a_{k-1}=S_{k-1}$. Thus, there does not exist $S \subset A$ such that $\sigma(S)=S_{k-1}+1$. Therefore,
$$
S_{k}=S_{k-1}+a_{k} \leqslant 2 S_{k-1}+1,
$$

By the problem statement, $S_{1}=a_{1}=1$. Thus, by (1) and induction, we can deduce that
$$
S_{k} \leqslant 2^{k}-1(1 \leqslant k \leqslant m),
$$

From $1500 \leqslant S_{m} \leqslant 2^{m}-1$, we get $m=11$.
From (1), we have
$$
\frac{S_{k+1}-1}{2} \leqslant S_{k} \leqslant 2^{k}-1 \quad(k=1,2, \cdots, 10).
$$

Since $S_{k}$ is a natural number, we have
$$
\left\lceil\frac{S_{k+1}-1}{2}\right\rceil \leqslant S_{k} \leqslant 2^{k}-1,
$$

where $\lceil x\rceil$ denotes the smallest integer greater than or equal to $x$. In particular,
$$
S_{10} \geqslant\left\lceil\frac{S_{11}-1}{2}\right\rceil \geqslant\left\lceil\frac{1500-1}{2}\right\rceil=750,
$$

Also, $S_{8} \leqslant 2^{8}-1=255$, so
$$
2 a_{10}>a_{9}+a_{10}=S_{10}-S_{8} \geqslant 495.
$$

Thus, $a_{10} \geqslant 248$.
On the other hand, we can construct a set $A$ that meets the requirements and has $a_{10}=248$. Specifically, take
$$
A=\{1,2,4,8,16,32,64,128,247,248,750\}.
$$

Since $255=2^{7}+2^{6}+\cdots+2+2^{0}$,
using binary representation, we can see that for $n \leqslant 255$, there exists $S \subset\{1,2,4, \cdots, 128\}$ such that $\sigma(S)=n$. Therefore, for $n \leqslant 255+247=502$, there exists $S \subset\{1,2, 4, \cdots, 128,247\}$ such that $\sigma(S)=n$; for $n \leqslant 502+248=750$, there exists $S \subset \{1,2, \cdots, 128,247,248\}$ such that $\sigma(S)=n$; for $n \leqslant 750+750=1500$, there exists $S \subset A$ such that $\sigma(S)=n$.