Ukrajina 2008

It's clear that the number $2000n^2$ does not influence the answer therefore the sought digits are the same for numbers $A=n^3+2009n^2+27n$ and $B=n^3+9n^2+27n$.

As the number $(B+27)$ equals $(n+3{)}^3$ (being the cube of the natural number) and ends in 0, this number should end in 000. Thus $B=\overline{X000}-27=\overline{Y073}$ where $X,Y$ are some natural numbers. Therefore the last three digits of the number are 073.