jmc

Let $x$ be the positive integer. The problem implies that $(3x{)}^2-x=2010$, or re-arranging, that $9x^2-x-2010=0$. Suppose this factors as $9x^2-x-2010=(ax+b)(cx+d)=ac{x}^2+(bc+ad)x+bd$. We can factor $2010=2\cdot 3\cdot 5\cdot 67$. If both $a$ and $c$ are divisible by $3$, then $bc+ad$ is also divisible by $3$, which is not the case. Thus, one of $a$ and $c$ is equal to $9$, and the other is equal to $1$; we will choose $a=9$. Then $b+9d=-1$ and $bd=2010$; after a bit of experimentation, we find that $b=2\cdot 67,d=3\cdot 5$ works. Thus, $$9x^2-x-2010=(9x+134)(x-15)=0,$$ and since $x$ is a positive integer, then $x=\boxed{15}$.