2003's math olympiad

math1

Can one find 4004 positive integers such that the sum of any 2003 of them is
not divisible by 2003?

Proof?:

Suppose we have 4004 positive integers (lets call them the set W), so that 2002 of them (the set T1) are of the type n∙2003 + 1 (where n is different for each of them) and 2002 of them (the set T2) are of the type m∙2003 + 2 (where m is different for each of them).

When we choose 2003 numbers from W, there will be an even number of T1s and an odd number of T2s or an odd number of T1s and an even of T2s.
So the two cases are:
Case 1:
2a(X∙2003 + 1) + (2b+1)(X∙2003 + 2)
2a∙X∙2003 + 2a + (2b+1)∙Y∙2003 + 2(2b+1)

2a∙X∙2003 and (2b+1)∙Y∙2003 are both divisable by 2003 but 2a + 4b + 2 is even and therefore is not equal to 2003 (and can't be 4006 because the biggest value it can has is 2*2002 + 1 = 4005) therefore the whole expression is not divisable to 2003.

Case 2:
(2a+1)(X∙2003 + 1) + 2b(X∙2003 + 2)
(2a+1)∙X∙2003 + 2a+1 + 2b∙Y∙2003 + 4b

2a+1 + 4b is odd so the whole expression can be divisible by 2003 if and only if 2a+1 + 4b is divisible by 2003. But 2a+1 + 2b = 2003 and 2a+1 + 2b = 2a+1 + 4b is possible only if b = 0, but this is impossible because this will mean that T1 has size of 2003 and that's not true (T1 has size of 2002)