11. Let $\sigma(\mathrm{S})$ denote the sum of all elements in a non-empty set of integers $S$. 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.
11. Solution: Let $S_{k}=a_{1}+a_{2}+\cdots+a_{k}(1 \leqslant k \leqslant 11)$, if $a_{k}>S_{k-1}+1$, then there does not exist $S \subset A$ such that $\sigma(S)=S_{k-1}+1$, so $S_{k}=S_{k-1}+a_{k} \leqslant 2 S_{k-1}+1$. (1)
From the given condition, $S_{1}=a_{1}=1$, thus by (1) and induction, it is easy to get $S_{k} \leqslant 2^{k}-1(1 \leqslant k \leqslant 11)$,
If $S_{10}<750$, then $a_{11} \leqslant 750$ (otherwise 750 cannot be represented by $\sigma(S)$), $S_{11}=S_{10}+a_{11}<1500$, so $S_{10} \geqslant 750$.
Also, $S_{8} \leqslant 2^{8}-1=255$, thus $2 a_{10} \geqslant a_{9}+a_{10}=S_{10}-S_{8} \geqslant 495$, so $a_{10} \geqslant 248$.
On the other hand, let $A=\{1,2,4,8,16,32,64,128,247,248,750\}$.
When $n \leqslant 255=2^{7}+2^{6}+\cdots+2^{1}+2^{0}$, we can find $S \subset\{1,2,4, \cdots, 128\}$ such that $\sigma(S)=n$.
When $n \leqslant 255+247=502$, there exists $S \subset\{1,2,4, \cdots, 128,247\}$ such that $\sigma(S)=n$.
When $n \leqslant 502+248=750$, there exists $S \subset\{1,2,4, \cdots, 128,247,248\}$ such that $\sigma(S)=n$.
When $n \leqslant 750+750=1500$, there exists $S \subset A$ such that $\sigma(S)=n$. Therefore, the minimum value of $a_{10}$ is 248.