smc

Consider the graph of $f(x)=|x-a|+|x-b|+|x-c|$. When $x<a$, the slope is $-3$. When $a<x<b$, the slope is $-1$. When $b<x<c$, the slope is $1$. When $c<x$, the slope is $3$. Setting $x=b$ gives $y=|b-a|+|b-b|+|b-c|=c-a$, so $(b,c-a)$ is a point on $f(x)$. In fact, it is the minimum of $f(x)$ considering the slope of lines to the left and right of $(b,c-a)$. Thus, graphing this will produce a figure that looks like a cup: From the graph, it is clear that $f(x)$ and $2x+y=2003$ have one intersection point if and only if they intersect at $x=a$. Since the line where $a<x<b$ has slope $-1$, the positive difference in $y$-coordinates from $x=a$ to $x=b$ must be $b-a$. Together with the fact that $(b,c-a)$ is on $f(x)$, we see that $P=(a,c-a+b-a)$. Since this point is on $x=a$, the only intersection point with $2x+y=2003$, we have $2\cdot a+(b+c-2a)=2003⟹b+c=2003$. As $c>b$, the smallest possible value of $c$ occurs when $b=1001$ and $c=1002$. This is indeed a solution as $a=1000$ puts $P$ on $y=2003-2x$, and thus the answer is $\boxed{1002}$. This indeed works for the two right segments of slope $1$ and $3$. We already know that the minimum is achieved between slopes $-3$ and $-1$ with $b+c=2003$: $$2003-2x=-a-b+c+x⟶3x\ne a+b-c+2003\{b<x<c\}\to (3b,3c)\ne a+2b\to b>a\text{ (true)}$$ $$2003-2x=-a-b-c+3x⟶5x\ne a+b+c+2003\{x>c\}\to (5c,+5\infty )\ne a+2b+2c\to 3c>a+2b\text{ (true)}$$ Indeed, within the restricted domain of $x$ in each segment, these inequalities prove to be unequal everywhere. So $y=2003-2x$ is strictly below $y=|x-a|+|x-b|+|x-c|$ at these domains.