jmc

By the Euclidean algorithm, we have $$\begin{array}{rl}& \text{gcd}(3a^2+19a+30,a^2+6a+9)\\ & =\text{gcd}(3a^2+19a+30,3a^2+19a+30-3(a^2+6a+9))\\ & =\text{gcd}(3a^2+19a+30,a+3)\\ & =a+3,\end{array}$$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 $\boxed{3}$.