(9) Given a sequence of positive terms $\left\{a_{n}\right\}$ satisfies: (1) $a_{1}=2012$; (2) $a_{2}, a_{3}$ are integers; (3) the sequence $\left\{n a_{n}-n^{2}\right\}$ is a geometric sequence with a common ratio no greater than 10. Find the general term formula of the sequence $\left\{a_{n}\right\}$.
9. From condition (3), we know that $n a_{n}-n^{2}=c \cdot q^{n-1}$, where $c, q>0$, thus
$$
a_{n}=\frac{c \cdot q^{n-1}}{n}+n, \quad n=1,2, \cdots \text {. }
$$

From condition (1), we get $c=2011$, hence
$$
a_{n}=\frac{2011 q^{n-1}}{n}+n, \quad n=1,2, \cdots \text {. }
$$

Since $a_{2}=\frac{2011 q}{2}+2$ is an integer, $\frac{2011 q}{2}$ must be an integer, so $q$ can only be a fraction. Let's assume $q=\frac{k}{m}$, where $k$ and $m$ are coprime. Noting that 2011 is a prime number, $m$ can only be 1 or 2011, and $k$ must be even.

Similarly, from $a_{3}=\frac{2011\left(\frac{k}{m}\right)^{2}}{3}+3$ being an integer, we know that $\frac{2011 \cdot \frac{k^{2}}{m^{2}}}{3}$ is an integer, so $m$ can only be 1 and $k$ must be a multiple of 3, thus $q=k$ must be a multiple of 6. Since $q$ is no greater than 10, $q=6$. Therefore, the general term formula for the sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=\frac{2011 \cdot 6^{n-1}}{n}+n, \quad n=1,2, \cdots \text {. }
$$