Answer: $ t = 4$ Solution: Consider the equation modulo $ 9$. Note that the cube of every integer is $ -1,0,$ or $ 1\pmod 9$. However, $ 2002^{2002}\equiv 4^{4}\pmod 9\equiv 256\pmod 9\equiv 4\pmod 9$. So, $ x_{1}^{3}+x_{2}^{3}+...+x_{t}^{3}\equiv 4\pmod 9$ but as $ |x_{i}\pmod 9|\leq 1$ we must have $ t\geq 4$. Now, if we take $ x_{1}= x_{2}= 10*2002^{667}, x_{3}= x_{4}= 2002^{667}$ then $ x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}= 2002^{667*3}(10^{3}+10^{3}+1^{1}+1^{3}) = 2002^{2001}*(2002) = 2002^{2002}$ So, for $ t = 4$ there exist the required integers $ x_{1}, x_{2},...,x_{t}$ and as $ t\geq 4$ then $ 4$ is the smallest positive integer satisfying the required conditions.

very easy and nice t=4

hints

imo shortlist 2002 N1