Given that

$m=\left(9n^2+54n+80\right)\left(9n^2+45n+54\right)\left(9n^2+36n+35\right)$

$\Rightarrow m = \left(9n^2+54n+81-1\right)9\left(\left(n+2\right)\left(n+3\right)\right)\left(9n^2+36n+36-1\right)$

$\Rightarrow m = \left[{\left(3n+9\right)}^2-1\right]9\left[\left(n+2\right)\left(n+3\right)\right]\left[{\left(3n+6\right)}^2-1\right]$

$\Rightarrow m = \left(3n+6\right)\left(3n+9\right)\left(3n+8\right)\left(3n+10\right)\left(3n+5\right)\left(3n+7\right)$

product of six consecutive numbers is always divisible by $6!$

So greatest positive integer which divides $m$ is $720$.