The empty set, $\varnothing$. If $m$ is relatively prime to $n$ and $m<n$, then $n-m$ must likewise be relatively prime to $n$, and these are distinct for $n>2$ since $n / 2, n$ are not relatively prime. Therefore, for all $n>2, \varphi(n)$ must be even. $2002^{2}-1$ is odd, and $\varphi(2)=1 \neq 2002^{2}-1$, so no numbers $n$ fulfill the equation.