jmc

Let the tens digit of the two-digit integer be $a$ and let its units digit be $b$. The equation $$10a+b=2(a+b)$$ is given. Distributing on the right-hand side and subtracting $2a+b$ from both sides gives $8a=b$. Since $8a>9$ for any digit $a>1$, we have $a=1$, $b=8$, and $10a+b=\boxed{18}$.