jmc

Call the three integers $x-2$, $x$, and $x+2$. We know that $(x-2)x(x+2)=20(x-2+x+x+2)$, or $(x^2-4)x=20(3x)$. Canceling an $x$ on either side gives $(x^2-4)=60$, so $x^2=64$. Thus, $x=8$, (since the numbers are positive), so the sum of the three integers is $3\cdot 8=\boxed{24}$.