If $f(1)=1$ and $f(1)+f(2)+\cdots+f(n)=n^{2} f(n)$ for every integer $n \geq 2$, evaluate $f(2008)$.
$n^{2} f(n)-f(n)=f(1)+f(2)+\cdots+f(n-1)=(n-1)^{2} f(n-1)$ hence $f(n)=\frac{(n-1)^{2}}{n^{2}-1} f(n-1)=\frac{n-1}{n+1} f(n-1)$. Thus $f(2008)=\frac{2007}{2009} f(2007)=\frac{2007}{2009} \cdot \frac{2006}{2008} f(2006)=\frac{2007}{2009} \cdot \frac{2006}{2008} \cdot \frac{2005}{2007} f(2005)=\cdots=\frac{2007 !}{2009 \cdot 2008 \cdots 4 \cdot 3} f(1)=\frac{2}{2009 \cdot 2008}$.