9. (14 points) Given a positive sequence $\left\{a_{n}\right\}$ 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 $\left\{a_{n}\right\}$.
From condition (3), we know that $n a_{n}-n^{2}=c q^{n-1}$.
Thus, $a_{n}=\frac{c q^{n-1}}{n}+n \quad (n=1,2, \cdots)$.
From condition (1), we get $c=2011$. Therefore,
$$
a_{n}=\frac{2011 q^{n-1}}{n}+n \quad (n=1,2, \cdots).
$$

Since $a_{2}=\frac{2011 q}{2}+2$ is an integer, $\frac{2011 q}{2}$ must be an integer.
Assume $q=\frac{k}{m} \quad ((k, m)=1)$.
Note that 2011 is a prime number. Hence, $m$ can only be 1 or 2011, and $k$ must be even.

Since $a_{3}=\frac{2011\left(\frac{k}{m}\right)^{2}}{3}+3$ is an integer, $\frac{2011 k^{2}}{3 m^{2}}$ must be an integer.
Thus, $m$ can only be 1 and $3 \mid k$.
Therefore, $q=k$ is a multiple of 6.
Since $q$ is no greater than 10, $q=6$.
Hence, $a_{n}=\frac{2011 \times 6^{n-1}}{n}+n \quad (n=1,2, \cdots)$.