By sum-to-product, \[\cos n \theta + \cos ((n - 2) \theta) = 2 \cos \theta \cos ((n - 1) \theta),\]or \[\cos n \theta = 2 \cos \theta \cos ((n - 1) \theta) - \cos ((n - 2) \theta)\]for all $ n \ge 2 $. In particular, for $ n = 2, $ \[\cos 2 \theta = 2 \cos^2 \theta - 1,\]and for $ n = 3, $ \begin{align*} \cos 3 \theta &= 2 \cos \theta \cos 2 \theta - \cos \theta \\ &= \cos \theta (2 \cos 2 \theta - 1).\end{align*}Suppose $ \cos \theta $ is irrational, and $ \cos 2 \theta $ and $ \cos 3 \theta $ are rational. Then $ 2 \cos 2 \theta - 1 $ is also rational, so we have a rational number that is the product of an irrational number and a rational number. The only way this can occur is if both rational numbers are 0. Thus, $ 2 \cos 2 \theta - 1 = 0 $. Then \[2 (2 \cos^2 \theta - 1) - 1 = 0,\]so $ \cos^2 \theta = \frac{3}{4} $. Hence, $ \cos \theta = \pm \frac{\sqrt{3}}{2} $. If $ \cos \theta = \frac{\sqrt{3}}{2}, $ then \begin{align*} \cos 2 \theta &= 2 \cos^2 \theta - 1 = \frac{1}{2}, \\ \cos 3 \theta &= 2 \cos \theta \cos 2 \theta - \cos \theta = 0, \\ \cos 4 \theta &= 2 \cos \theta \cos 3 \theta - \cos 2 \theta = -\frac{1}{2}, \\ \cos 5 \theta &= 2 \cos \theta \cos 4 \theta - \cos 3 \theta = -\frac{\sqrt{3}}{2}, \end{align*}so the largest possible value of $ n $ is 4. Similarly, if $ \cos \theta = -\frac{\sqrt{3}}{2}, $ then \begin{align*} \cos 2 \theta &= 2 \cos^2 \theta - 1 = \frac{1}{2}, \\ \cos 3 \theta &= 2 \cos \theta \cos 2 \theta - \cos \theta = 0, \\ \cos 4 \theta &= 2 \cos \theta \cos 3 \theta - \cos 2 \theta = -\frac{1}{2}, \\ \cos 5 \theta &= 2 \cos \theta \cos 4 \theta - \cos 3 \theta = \frac{\sqrt{3}}{2}, \end{align*}so again the largest possible value of $ n $ is 4. Therefore, the largest possible value of $ n $ is $ \boxed{4} $.