Twelve Distinct Positive Integers

Consider the ${12 \choose 2} = 66$ possible differences $b-a$, where $b>a$ are the elements of $S$.

Note that if $a, b, c, d \in S$ are pairwise distinct and $a-c = d-b$, then $a+b = c+d$.

Also, if $a,b,c \in S$ are all distinct and $c-a = b-c$, then $c = (a+b)/2$ so $c$ is not the largest or smallest element of $S$.

Next, if $c-a = b -c$ and $c-d = e-c$, then $a+b = 2c = d+e$. So for each $c$, that is not the largest or smallest element of $S$, there exists at most one pair of $a, b$ that can be found such that $c-a=b-c$. Because if not, then we will have $a + b = d +e$.

Hence, there are at most $10$ duplicates for each $c$ that is not the largest or smallest element of $S$. Therefore, there are at least $66 - 10 = 56$ distinct differences.

Hence, the largest element of $S$ is greater than 56.