The discussion proves that the expression $1 \times 3 \times 5 \times \ldots \times 1999 \times 2001 + 2 \times 4 \times 6 \times \ldots \times 2000 \times 2002$ is a multiple of 2003. The second part of the expression can be rewritten as $(2003-1)(2003-3)\ldots(2003-2001)$, demonstrating that every term in the expansion contains 2003, except for the last term, which is $(1 \times 3 \times 5 \ldots \times 2001)(-1)^{1001}$. This last term is negative and cancels out the first part of the expression, confirming that 2003 divides every term, thus proving the statement definitively.
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