jmc

There are different ways to approach this problem, and I'll start with the different factor of the numbers of the set $\{1,2,3,4\}$. $1$ has factor $1$. $2$ has factors $1$ and $2$ $3$ has factors $1$ and $3$ $4$ has factors $1$, $2$, and $4$. From here, we note that even though all numbers have the factor $1$, only $4$ has another factor other than $1$ in the set (ie. $2$) We could therefore have one fraction be $\frac{4}{2}$ and another $\frac{3}{1}$. The sum of the numerators is $4+3=7=\boxed{7}$