$11 \cdot 6$ For a non-empty set $S$ of integers, let $\sigma(S)$ denote the sum of all elements in $S$. 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 possible value of $a_{10}$.
(21st United States of America Mathematical Olympiad, 1992)
[Solution] Let $S_{k}=\sum_{i=1}^{k} a_{i}$, then we have
$$
a_{k+1} \leqslant S_{k}+1, \quad k=1,2, \cdots, 10 .
$$

In fact, if there exists $j$ such that
$$
a_{j+1}>S_{j}+1,
$$

then for $n=S_{j}+1$, since
$$
a_{1}+a_{2}+\cdots+a_{j}=S_{j}S_{j}+1=n \quad(t>j) .
$$

therefore, there does not exist $S \subset A$ such that $\sigma(S)=n$.
It is easy to see that $a_{1}=1$, thus we can obtain
$$
\begin{array}{l}
a_{2} \leqslant 2, \quad a_{3} \leqslant 4, \quad a_{4} \leqslant 8, \quad a_{5} \leqslant 16, \\
a_{6} \leqslant 32, \quad a_{7} \leqslant 64, \quad a_{8} \leqslant 128 .
\end{array}
$$

If $a_{10} \leqslant 247$, then $a_{9} \leqslant 246$, hence
$$
\begin{aligned}
a_{11} & \leqslant S_{10}+1 \\
& =(1+2+4+\cdots+128)+246+247+1 \\
& =749 .
\end{aligned}
$$

Thus
$$
S_{11}=S_{10}+a_{11} \leqslant 1497
$$

This is impossible.
Therefore, $a_{10} \geqslant 248$.
Moreover, when $a_{1}=1, a_{2}=2, a_{3}=4, a_{4}=8, a_{5}=16, a_{6}=32$, $a_{7}=64, \quad a_{8}=128, \quad a_{9}=247, \quad a_{10}=248, \quad a_{11}=751$, the set $A=\left\{a_{1}, a_{2}, \cdots, a_{11}\right\}$ satisfies the problem's requirements.
Indeed, $1,2,4,8, \cdots, 128$ can represent all integers from 1 to $2^{7}+2^{6}+\cdots+2+1=255$, which can be derived from binary representation.
$$
1,2,4, \cdots, 128,247,248 \text { can represent all integers from } 248 \text { to } 247+248+255=750 \text { }
$$

Thus, $1,2, \cdots, 128,247,248,751$ can represent all integers from 1 to 1500.