Solution 1. Consider a positive integer $n$ with $n!+1 \mid(2012 n)$ !. It is well-known that for arbitrary nonnegative integers $a_{1}, \ldots, a_{k}$, the number $\left(a_{1}+\ldots+a_{k}\right)$ ! is divisible by $a_{1}!\cdot \ldots \cdot a_{k}!$. (The number of sequences consisting of $a_{1}$ digits $1, \ldots, a_{k}$ digits $k$, is $\frac{\left(a_{1}+\ldots+a_{k}\right)!}{a_{1}!\ldots \ldots a_{k}!}$.) In particular, $(n!)^{2012}$ divides $(2012 n)!$. Since $n!+1$ is co-prime with $(n!)^{2012}$, their product $(n!+1)(n!)^{2012}$ also divides $(2012 n)$ !, and therefore $$(n!+1) \cdot(n!)^{2012} \leq(2012 n)!$$ By the known inequalities $\left(\frac{n+1}{e}\right)^{n}<n!\leq n^{n}$, we get $$\left(\frac{n}{e}\right)^{2013 n}<(n!)^{2013}<(n!+1) \cdot(n!)^{2012} \leq(2012 n)!<(2012 n)^{2012 n}$$ Therefore, $n<2012^{2012} e^{2013}$. Therefore, there are only finitely many such integers $n$. Solution 2. Assume that $n>2012$ is an integer with $n!+1 \mid(2012 n)$ !. Notice that all prime divisors of $n!+1$ are greater than $n$, and all prime divisors of (2012n)! are smaller than $2012 n$. Consider a prime $p$ with $n<p<2012 n$. Among $1,2, \ldots, 2012 n$ there are $\left[\frac{2012 n}{p}\right]<2012$ numbers divisible by $p$; by $p^{2}>n^{2}>2012 n$, none of them is divisible by $p^{2}$. Therefore, the exponent of $p$ in the prime factorization of $(2012 n)$ ! is at most 2011. Hence, $n!+1=\operatorname{gcd}(n!+1,(2012 n)!)<\prod_{n<p<2012 p} p^{2011}$. Applying the inequality $\prod_{p \leq X} p<4^{X}$, $$n!<\prod_{n<p<2012 p} p^{2011}<\left(\prod_{p<2012 n} p\right)^{2011}<\left(4^{2012 n}\right)^{2011}=\left(4^{2012 \cdot 2011}\right)^{n}$$ Again, we have a factorial on the left-and side and a geometric progression on the right-hand side.