(N1 2002) Find the smallest integer $t$ such that there exist strictly positive integers $x_{1}, \ldots, x_{t}$, satisfying $x_{1}^{3}+\ldots+x_{t}^{3}=2002^{2002}$.
Let's show that $t \geq 4$. We are looking for an integer $n$ such that $3 \mid \varphi(n)$, so that cubes take on only a few values modulo $n$. This is the case for 9, for which cubes can only be 0, 1, or -1. We observe that $2002^{2002} \equiv 4 \pmod{9}$, so we have $t \geq 4$. It suffices to find $x_{1}, x_{2}, x_{3}, x_{4}$ that work to conclude. We have $2002 = 3 \cdot 667 + 1$, so $2002^{2002} = 2002 \cdot (2002^{667})^3$. It suffices to show that 2002 can be written as the sum of 4 cubes, which is easy to notice: $2002 = 1000 + 1000 + 1 + 1$.