Ukrajina 2008

It's clear that the number $2000n$ does not influence the answer, which implies that the sought digits will be the same for numbers $A=n^2+2008n$ and $B=n^2+8n^2$. As number $(B+16)$ equals $(n+4{)}^2$ (being the square of the natural number) and ends in $0$, this number should end in $00$. Thus $B=\overline{X00}-16=\overline{Y84}$ where $X,Y$ are some natural numbers. Therefore the last two digits of the number are $84$.