If $a$ is the smallest element of such a set, then $13-a$ is the largest element, and for the remaining elements we may choose any (or none) of the $12-2 a$ elements $a+1, a+2, \ldots,(13-a)-1$. Thus there are $2^{12-2 a}$ such sets whose smallest element is $a$. Also, $13-a \geq a$ clearly implies $a<7$. Summing over all $a=1,2, \ldots, 6$, we get a total of $$2^{10}+2^{8}+2^{6}+\cdots+2^{0}=4^{5}+4^{4}+\cdots+4^{0}=\left(4^{6}-1\right) /(4-1)=4095 / 3=1365$$ possible sets.