We can rewrite $2010$ as $2\cdot 3\cdot 5\cdot 67$. In order for $n^k-n$ to be a multiple of $2010$, it must be a multiple of each of these primes meaning that: \[n^k-n\equiv 0\pmod{2,3,5,67}.\] For modulo $2$, if $n$ is odd, $n^k-n$ is even, and if $n$ is even, $n^k-n$ is even so that part does not matter. If $n$ is congruent to any one of these other primes, $n^k-n$ is congruent to $0$ modulo that prime which means $k$ can be anything for that with no restrictions. Let us say this is not the case since we are looking at all positive integers. This means that: \[n^k\equiv n\pmod{3,5,67}\]\[n^{k-1}\equiv 1\pmod{3,5,67}.\]By FLT, $n^{p-1}\equiv 1\pmod{p}$. This means that $k-1$ is sometimes multiple of $p-1$ (it does not always have to be but it will for some values of $n$ and since we are considering all values of $n$, we have to assume this) getting: \[k-1\equiv 0\pmod{2,4,66},\]\[k\equiv 1\pmod{2,4,66}.\] By CRT, $k\equiv 1\pmod{132}$ which means the smallest possible value for $k$ is $133$.

trk08 wrote: By FLT, $n^{p-1}\equiv 1\pmod{p}$. This means that $k-1$ is a multiple of $p-1$ No it does not: $3^5 \cong 1 \mod 11$ and $10 \nmid 5$