USAMO

Let's a $n$ - $p$ set be a set $Z$ such that $Z=\{a_1,a_2,\cdots,a_n\}$ , where $\forall i<n$ , $i\in \mathbb{Z}^+$ , $a_i<a_{i+1}$ , and for each $x\le p$ , $x\in \mathbb{Z}^+$ , $\exists Y\subseteq Z$ , $\sigma(Y)=x$ , $\nexists \sigma(Y)=p+1$ .
(For Example $\{1,2\}$ is a $2$ - $3$ set and $\{1,2,4,10\}$ is a $4$ - $8$ set)
Furthermore, let call a $n$ - $p$ set a $n$ - $p$ good set if $a_n\le p$ , and a $n$ - $p$ bad set if $a_n\ge p+1$ (note that $\nexists \sigma(Y)=p+1$ for any $n$ - $p$ set. Thus, we can ignore the case where $a_n=p+1$ ).
Furthermore, if you add any amount of elements to the end of a $n$ - $p$ bad set to form another $n$ - $p$ set (with a different $n$ ), it will stay as a $n$ - $p$ bad set because $a_{n+x}>a_{n}>p+1$ for any positive integer $x$ and $\nexists \sigma(Y)=p+1$ .
Lemma ) If $Z$ is a $n$ - $p$ set, $p\leq 2^n-1$ .
For $n=1$ , $p=0$ or $1$ because $a_1=1 \rightarrow p=1$ and $a_1\ne1\rightarrow p=0$ .
Assume that the lemma is true for some $n$ , then $2^n$ is not expressible with the $n$ - $p$ set. Thus, when we add an element to the end to from a $n+1$ - $r$ set, $a_{n+1}$ must be $\le p+1$ if we want $r>p$ because we need a way to express $p+1$ . Since $p+1$ is not expressible by the first $n$ elements, $p+1+a_{n+1}$ is not expressible by these $n+1$ elements. Thus, the new set is a $n+1$ - $r$ set, where $r\leq p+1+a_{n+1} \leq 2^{n+1}-1$
Lemma Proven
The answer to this question is $\max{(a_{10})}=248$ .
The following set is a $11$ - $1500$ set:
$\{1,2,4,8,16,32,64,128,247,248,750\}$
Note that the first 8 numbers are power of $2$ from $0$ to $7$ , and realize that any $8$ or less digit binary number is basically sum of a combination of the first $8$ elements in the set. Thus, $\exists Y\subseteq\{1,2,4,8,16,32,64,128\}$ , $\sigma(Y)=x \forall 1\le x\leq 255$ .
$248\le\sigma(y)+a_9\le502$ which implies that $\exists A\subseteq\{1,2,4,8,16,32,64,128,247\}$ , $\sigma(A)=x \forall 1\le x\leq 502$ .
Similarly $\exists B\subseteq\{1,2,4,8,16,32,64,128,247,248\}$ , $\sigma(A)=x \forall 1\le x\le750$ and $\exists C\subseteq\{1,2,4,8,16,32,64,128,247,248,750\}$ , $\sigma(A)=x \forall 1\leq x\leq 1500$ .
Thus, $\{1,2,4,8,16,32,64,128,247,248,750\}$ is a $11$ - $1500$ set.
Now, let's assume for contradiction that $\exists a_{10}\leq 247$ such that ${a_1, a_2, \dots, a_{11}}$ is a $11$ - $q$ set where $q\geq 1500$
${a_1, a_2, \dots a_8}$ is a $8$ - $a$ set where $a\leq 255$ (lemma).
$max{(a_9)}=a_{10}-1\leq 246$
Let ${a_1, a_2, \dots, a_{10}}$ be a $10$ - $b$ set where the first $8$ elements are the same as the previous set. Then, $256+a_9+a_{10}$ is not expressible as $\sigma(Y)$ . Thus, $b\leq 255+a_9+a_{10}\leq 748$ .
In order to create a $11$ - $d$ set with $d>748$ and the first $10$ elements being the ones on the previous set, $a_{11}\leq 749$ because we need to make $749$ expressible as $\sigma(Y)$ . Note that $b+1+a_{11}$ is not expressible, thus $d<b+1+a_{11}\leq 1498$ .
Done but not elegant...