jmc

Let the two integers be $a$ and $b$. Then, $\gcd (a,b)=3$ and, without loss of generality, let $\mathrm{lcm}[a,b]=12a$. Multiplying the two equations yields that $\mathrm{lcm}[a,b]\cdot \gcd (a,b)=36a$. Using the identity that $ab=\mathrm{lcm}[a,b]\cdot \gcd (a,b)$, it follows that $ab=36a$, and so $b=36$.

Since $\gcd (a,b)=3$, we know $a$ is divisible by 3. However, $a$ cannot be divisible by $3^2=9$, because if $a$ was divisible by 9, then $\gcd (a,b)$ would be divisible by 9 as well, since 36 is divisible by 9. This cannot occur since $\gcd (a,b)=3$. Similarly, $a$ cannot be divisible by 2, because if $a$ were divisible by 2, then $\gcd (a,b)$ would be divisible by 2 as well, since 36 is divisible by 2.

In summary, $a$ is a multiple of 3, but not 9, and $a$ is not divisible by 2. The largest such number less than 100 is 93. We can verify that $\mathrm{lcm}[93,36]=1116=12\cdot 93$, so the largest possible sum of $a+b$ is $36+93=\boxed{129}$.