The equation implies that $n$ divides 2010. Therefore $n$ must be a positive divisor of 2010. Since 2010= 2*3*5*67 then $n \in (1,2,3,5,6,10,15,30,67,134,201,335,402,670,1005,2010)$ Plugging all this possible values of $n$ in the ecuation we find that the only number that works is $n=402$. $402(S(402)-1)= 402*5=2010$

Since $S(n)\equiv n\pmod9$ we have $n(n-1)\equiv 3\pmod9$, so $n\equiv 4,6\pmod 9$. We also have $n(n-1)\ge n(S(n)-1)=2010$ so $n>30$. Thus $n$ is a divisor of $2010$ which is greater than 30, which means $67|n$. But $S(n)\ge4$, so $n\le\frac{2010}{4-1}=670$. Thus $n\in\{67,134,201,335,402\}$. Since $n\equiv 4,6\pmod9$ then $n\in\{67,402\}$. After checking, we get that only 402 satisfies the given equation.