imc

Let $m$ be the number of main courses the restaurant serves, so $2m$ is the number of appetizers. Then the number of dinner combinations is $2m\times m\times 3=6m^2$. Since the customer wants to eat a different dinner in all $365$ days of $2003$, we must have $$\begin{array}{rl}6m^2& \ge 365\\ m^2& \ge 60.83\dots .\end{array}$$ Also, year 2003 is not a leap year, because 2003 divided by 4 does not equal an integer. The smallest integer value that satisfies this is $\boxed{8}$.