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