jmc

Let the integers be $a$ and $b$. Then $ab=144$ and $$\frac{\mathrm{lcm}[a,b]}{\gcd (a,b)}=9.$$The identity $ab=\gcd (a,b)\cdot \mathrm{lcm}[a,b]$ yields that $$ab=\gcd (a,b)\cdot \mathrm{lcm}[a,b]=144.$$Multiplying the two equations above yields that $(\mathrm{lcm}[a,b]{)}^2=9\cdot 144=36^2$, so $\mathrm{lcm}[a,b]=36$. Then $\gcd (a,b)=144/36=4$.

Since $\gcd (a,b)=4$ is a divisor of both $a$ and $b$, $a$ must have at least two factors of 2, and $b$ must have at least two factors of 2. Therefore, their product $ab$ has at least four factors of 2. But $ab=144=2^4\cdot 3^2$, which has exactly four factors of 2, so both $a$ and $b$ have exactly two factors of 2.

Since $ab=2^4\cdot 3^2$, the only primes that can divide $a$ and $b$ are 2 and 3. Let $a=2^2\cdot 3^u$ and let $b=2^2\cdot 3^v$. Then $\gcd (a,b)=2^2\cdot 3^{\min \{u,v\}}$. But $\gcd (a,b)=4=2^2\cdot 3^0$, so $\min \{u,v\}=0$, which means that either $u=0$ or $v=0$.

Hence, one of the numbers $a$ and $b$ must be 4, and the other number must be $144/4=36$. Therefore, the sum of the numbers is $4+36=\boxed{40}$.