imc

The terms in the first sequence are defined by $n\equiv 1(\mathrm{mod}3)$, while the terms in the second sequence are defined by $n\equiv 2(\mathrm{mod}7).$ We seek to find the solutions to this system of modular congruences. Letting $n=3a+1=7b+2$ for nonnegative integers $a,b$ we solve the congruence $$\begin{array}{rl}3a+1& \equiv 7b+2(\mathrm{mod}3),\\ 1& \equiv b+2(\mathrm{mod}3),\\ b& \equiv 2(\mathrm{mod}3).\end{array}$$ For nonnegative integer $c,$ we have that $b=3c+2$. Substituting back into our original equation, we have $n=21c+16,$ or $n\equiv 16(\mathrm{mod}21).$ Now we have to find the largest term in the smaller sequence (first sequence), which is $2003\cdot 3+1=6010.$ Dividing $\frac{6010}{21}$ gives that there are $286$ terms in common between the sequences. By PIE, the answer is just $2004+2004-286=\boxed{3722}.$