For $2 \leq i \leq 6$, we claim that $a_{1} \equiv \ldots \equiv a_{7}(\bmod i)$. This is because if we consider any $i-1$ of the 7 numbers, the other $7-(i-1)=8-i$ of them must all be equal modulo $i$, because we want the sum of all subsets of size $i$ to be a multiple of $i$. However, $8-i \geq 2$, and this argument applies to any $8-i$ of the 7 integers, so in fact all of them must be equal modulo $i$. We now have that all of the integers are equivalent modulo all of $2, \ldots, 6$, so they are equivalent modulo 60, their least common multiple. Therefore, if the smallest integer is $k$, then the other 6 integers must be at least $k+60, k+60 \cdot 2, \ldots, k+60 \cdot 6$. This means the sum is $7k+60 \cdot 21 \geq 7+60 \cdot 21=1267$. 1267 is achievable with $\{1,1+60, \ldots, 1+60 \cdot 6\}$, so it is the answer.