jmc

Since the system has exactly one solution, the graphs of the two equations must intersect at exactly one point. If $x<a$, the equation $y=|x-a|+|x-b|+|x-c|$ is equivalent to $y=-3x+(a+b+c)$. By similar calculations we obtain

$$y=\{\begin{array}{ll}-3x+(a+b+c),& \text{if }x<a\\ -x+(-a+b+c),& \text{if }a\le x<b\\ x+(-a-b+c),& \text{if }b\le x<c\\ 3x+(-a-b-c),& \text{if }c\le x.\end{array}$$Thus the graph consists of four lines with slopes $-3$, $-1$, 1, and 3, and it has corners at $(a,b+c-2a)$, $(b,c-a)$, and $(c,2c-a-b)$.

On the other hand, the graph of $2x+y=2003$ is a line whose slope is $-2$. If the graphs intersect at exactly one point, that point must be $(a,b+c-2a).$ Therefore

$2003=2a+(b+c-2a)=b+c.$

Since $b<c$, the minimum value of $c$ is $\boxed{1002}$.