a) since a sequence of 2003 consecutive integers may be written as $x+1, x+2, ..., x+2003$, it's enough to consider the case where both sequences are $1, 2, ..., 2003$. the following arrangement shows that it's possible: $\left(\begin{array} {ccccccccc} 1 & 1003 & 2 & 1004 & 3 & 1005 & {...} & 2003 & 1002 \\ 1002 & 1 & 1003 & 2 & 1004 & 3 & {...} &1001&2003 \\ \end{array} \right)$ b) consider more generally the case of two sequences each of length $2k$. as in (a) we may assume both sequences are $1, 2, ..., 2k$. the sum of all the elements of both these sequences is $2\cdot \frac{2k(2k+1)}{2} = 2k(2k+1)$. on the other hand, assume they produced a set of $2k$ consecutive integers $m+1, m+2, ..., m+2k$. the sum of these numbers is $2km+k(2k+1) = k(2m+2k+1)$, and should equal the sum $2k(2k+1)$. but this is impossible because the latter has one more factor of 2 than the former.