Let $ \omega $ be a root of $ x^2 + x + 1 = 0, $ so $ \omega^2 + \omega + 1 = 0 $. Then by the factor theorem, $ x^{2n} + 1 + (x + 1)^{2n} $ is divisible by $ x^2 + x + 1 $ if and only if $ \omega^{2n} + 1 + (\omega + 1)^{2n} = 0 $. Since $ \omega + 1 = -\omega^2, $ \[\omega^{2n} + 1 + (\omega + 1)^{2n} = \omega^{2n} + 1 + (-\omega^2)^{2n} = \omega^{4n} + \omega^{2n} + 1.\]From the equation $ \omega^2 + \omega + 1 = 0, $ $ (\omega - 1)(\omega^2 + \omega + 1) = \omega^3 - 1, $ so $ \omega^3 = 1 $. We divide into the cases where $ n $ is of the form $ 3k, $ $ 3k + 1, $ and $ 3k + 2 $. If $ n = 3k, $ then \begin{align*} \omega^{4n} + \omega^{2n} + 1 &= \omega^{12k} + \omega^{6k} + 1 \\ &= (\omega^3)^{4k} + (\omega^3)^{2k} + 1 \\ &= 1 + 1 + 1 = 3.\end{align*}If $ n = 3k + 1, $ then \begin{align*} \omega^{4n} + \omega^{2n} + 1 &= \omega^{12k + 4} + \omega^{6k + 2} + 1 \\ &= (\omega^3)^{4k + 1} \omega + (\omega^3)^{2k} \omega^2 + 1 \\ &= \omega + \omega^2 + 1 = 0.\end{align*}If $ n = 3k + 2, $ then \begin{align*} \omega^{4n} + \omega^{2n} + 1 &= \omega^{12k + 8} + \omega^{6k + 4} + 1 \\ &= (\omega^3)^{4k + 2} \omega^2 + (\omega^3)^{2k + 1} \omega + 1 \\ &= \omega^2 + \omega + 1 = 0.\end{align*}Hence, $ x^{2n} + 1 + (x + 1)^{2n} $ is divisible by $ x^2 + x + 1 $ if and only if $ n $ is of the form $ 3k + 1 $ or $ 3k + 2, $ i.e. is not divisible by 3. In the interval $ 1 \le n \le 100, $ there are $ 100 - 33 = \boxed{67} $ such numbers.