jmc

Let $n=2003.$ Then the sides of the box are $n,$ $n+1,$ and $n(n+1)=n^2+n,$ so if $d$ is the length of the space diagonal of the box, then $$d^2=n^2+(n+1{)}^2+(n^2+n{)}^2=n^4+2n^3+3n^2+2n+1.$$Note that $n^4+2n^3+3n^2+2n+1=(n^2+n+1{)}^2,$ so $$d=n^2+n+1=\boxed{4014013}.$$