6. Let $\sigma(\mathrm{s})$ denote the sum of all elements in a non-empty set of integers $\mathrm{S}$. Let $A=\left\{a_{1}, a_{2}, \mathrm{~L} a_{11}\right\}$ be a set of positive integers, and $a_{1}<a_{2}<\mathrm{L} a_{11}$. If for each positive integer $n \leq 1500$, there exists a subset $\mathrm{S}$ of $\mathrm{A}$ such that $\sigma(S)=n$, then the minimum value of $a_{10}$ that satisfies the above requirement is $\qquad$
6.248

Analysis: It is easy to know that $a_{1}=1$. From the given, we have: $a_{1}+a_{2}+\mathrm{L}+a_{k-1}+1 \geq a_{k}$, otherwise, if $a_{1}+a_{2}+\mathrm{L}+a_{k-1}+1<a_{k}$, then $a_{1}+a_{2}+\mathrm{L}+a_{k-1}+1$ cannot be represented. Therefore, $S_{k}=S_{k-1}+a_{k} \leq 2 S_{k-1}+1$, which means $S_{k} \leq 2^{k}-1(1 \leq k \leq 11)$; if $S_{10}<750$, then $a_{11} \leq 750, S_{11}<1500$, so $S_{10} \geq 750$; also, $S_{8} \leq 2^{8}-1=255$, then $2 a_{10} \geq a_{9}+a_{10}=S_{10}-S_{8} \geq 495$, hence $a_{10} \geq 248$; on the other hand, we take $A=\{1,2,4,8,16,32,64,128,247,248,750\}$, it can be known that when $n \leq 255$, we can take $S \subset\{1,2,4,8,16,32,64,128\}$, such that $\sigma(S)=n$; when $n \leq 255+247=502$, we can take $S \subset\{1,2,4,8,16,32,64,128,247\}$, such that $\sigma(S)=n$; when $n \leq 502+248=750$, we can take $S \subset\{1,2,4,8,16,32,64,128,247,248\}$, such that $\sigma(S)=n$; when $n \leq 750+750=1500$, we can take $S \subset A$, such that $\sigma(S)=n$; therefore, the minimum value of $a_{10}$ is 248.