If $n$ is a composite number and doesn’t equal to a square of two, then $\frac{2008^p}{n}$ is an integer (we wish to prove it). Obviously it’s possible if only all prime divisors of $n$ are prime divisors of $2008$. If $n$ is a prime number then $(n-1)!\equiv -1 $ mod $n$, so $ord_n(2008)|2p$, but $p>n-1> ord_n(2008)$, hence $ord_n(2008)=2$, that is $n$ is a prime divisor of $2009$. If $n=4$, then $4$ doesn’t divide $3!$. Therefore, all such $n$ are either prime divisors of $2009$ or composite numbers such that any prime divisor of them divides $2008$.