Assume the contrary, namely $ \mathrm{arccos} \left( \frac{1}{\sqrt{2003}} \right) = \frac{\pi r}{n}$, for some $ r, n \in \mathbb{N}$, $ (r, n) = 1$ and $ r < \frac{n}{2}$. Taking the cosine of the both sides and, then, squaring yields $ \frac{1}{2003} = \cos^2 \frac{\pi r}{n} = \frac{1}{2} \cdot \left( \cos \frac{2 \pi r}{n} + 1 \right)$. Hence $ \cos \frac{2\pi r}{n} = - \frac{2001}{2003} \in \mathbb{Q}$, and by a well-known result $ n \in \{1, 2, 3, 4, 6\}$, however it is a check by hand that in no such case $ \cos \frac{2\pi r}{n} = - \frac{2001}{2003}$, the contradiction needed.