jmc

Call the two positive integers $x$ and $y$. Without loss of generality, assume $x>y$. We can write a system of equations to represent the information given in the problem: $$\begin{array}{rl}x-y& =2\\ x\cdot y& =120\end{array}$$ Solving for $x$ in the first equation yields $x=y+2$.

Substituting this into the second equation gives $(y+2)\cdot y=120$, or $y^2+2y-120=0$.

This quadratic equation factors into $(y+12)(y-10)=0$, so $y=10$.

Given $y$, we can solve for $x$ to get $x=12$, so $x+y=\boxed{22}$.