By the Euclidean algorithm, we have \begin{align*} &\text{gcd}(3a^2+19a+30,a^2+6a+9) \\ &\qquad= \text{gcd}(3a^2+19a+30,3a^2+19a+30-3(a^2+6a+9)) \\ &\qquad= \text{gcd}(3a^2+19a+30,a+3) \\ &\qquad= a+3, \end{align*}since the integer $ 3a^2+19a+30 $ is divisible by $ a+3 $ for all integers $ a $, as the factorization $ 3a^2+19a+30=(3a+10)(a+3) $ shows. Thus $ f(a)-a $ is equal to 3 for all positive integers $ a $, so its maximum value is $ \boxed{3} $.