2004 problem!

$\begin{aligned} S & = 1^{2003} + 2^{2003} + 3^{2003} + ... + 2003^{2003} \\ & = (1^{2003} + 2003^{2003}) + (2^{2003} + 2002^{2003}) + (3^{2003} + 2001^{2003}) + ... + (1001^{2003} + 1003^{2003}) + 1002^{2003} \\ & = \sum_{n=1}^{1001} \left(n^{2003} + (2004-n)^{2003}\right) + 1002^{2003} \\ & = \sum_{n=1}^{1001} \left[(n + (2004-n)) \left(n^{2002} - n^{1001}(2004-n)^{1001} + (2004-n)^{2002} \right) \right] + 1002^{2003} \\ & = 2004 \sum_{n=1}^{1001} \left(n^{2002} - n^{1001}(2004-n)^{1001} + (2004-n)^{2002} \right) + 1002^{2003} \end{aligned}$

$\begin{aligned} \implies S & \equiv 1002^{2003} \pmod{2004} \\ & \equiv (2 \cdot 3 \cdot 167)^{2003} \pmod{2^2 \cdot 3 \cdot 167} \\ & \equiv 1002(2^2 \cdot 3^2 \cdot 167^2)^{1001} \pmod{2^2 \cdot 3 \cdot 167} \\ & \equiv 1002(3 \cdot 167 \cdot 2004)^{1001} \pmod{2004} \\ & \equiv \boxed{0} \pmod{2004} \end{aligned}$

Pair the first and last term second and second to last term and so on. Each of these pairs will be of the form n^2003 +(-n)^2003 which is congruent to zero

$\begin{matrix} 1^{ 2003 }+2^{ 2003 }+3^{ 2003 }+\dots +2003^{ 2003 } & = & (1^{ 2003 }+2003^{ 2003 })+(2^{ 2003 }+2002^{ 2003 })+\dots +(1001^{ 2003 }+1003^{ 2003 })+1002^{ 2003 } \\ & = & (1^{ 2003 }+(-1)^{ 2003 })+(2^{ 2003 }+(-2)^{ 2003 })+\dots +(1001^{ 2003 }+(-1001)^{ 2003 })+1002^{ 2003 } \\ & = & 1002^{ 2003 }\equiv 0\quad ({ \bmod \text{ 2004} }) \end{matrix}$