jmc

Using the Euclidean Algorithm, we obtain $$\begin{array}{rl}\text{gcd}(2m,3n)& =\text{gcd}(2m-3n\cdot 16,3n)\\ & =\text{gcd}(2\cdot (24n+51)-48n,3n)\\ & =\text{gcd}(102,3n)\\ & \le 102.\end{array}$$Therefore, the largest possible value of $\text{gcd}(2m,3n)$ is $\boxed{102}$. For example, this is possible when $n=34\Rightarrow m=867$, as $3n=102$ and $2m=1734=102\cdot 17$.