I claim that for all $ n \in \mathbb{N}$, $ S(n) : = \sum_{k = 1}^n \tan^2 \frac {k\pi}{2n + 1} = n(2n + 1)$. First of all, denote $ t_k : = \frac {k\pi}{2n + 1}$, where $ k$ ranges from $ 1$ to a fixed $ n$, and $ m = 2n + 1$. Now, in the relation $ \frac {\cos(mx) + i\sin(mx)}{(\cos(x))^m} = \frac {(\cos(x) + i\sin(x))^m}{(\cos(x))^m} = (1 + i \tan(x))^m =$ $ \sum_{k, 2 | k} ( - 1)^{k/2} \binom{m}{k} \tan^k(x) + i \cdot \sum_{k, 2 | (k - 1)} ( - 1)^{(k - 1)/2} \binom{m}{k} \tan^k(x)$, equating the imaginary parts and rewriting the sum yield $ \frac {\sin(mx)}{(\cos(x))^m} = \tan(x) \cdot \sum_{k = 0}^n ( - 1)^k \binom{2n + 1}{2k + 1} (\tan^2 (x))^k$. Now assigning to $ x$ all the values of $ t_k$ causes the left-hand side to vanish, hence the distinct numbers $ \tan^2(t_k)$ are all the roots of $ P \equiv \sum_{k = 0}^n ( - 1)^k \binom{2n + 1}{2k + 1} X^k \in \mathbb{Z}[X]$. This shows that $ S(n)$ is equal, by Viete's relations, to $ - \frac {[X^{n - 1}] P}{[X^n] P} = n(2n + 1)$, which was to be proved (I denoted by $ [X^t] P$ the coefficient in $ P$ of the monomial of degree $ t$). Consequently the problem is not correct - the sum is odd iff $ n$ is odd. I strongly suggest changing the statement into either "find a closed form" or "prove that the sum is equal to [...]" - any others represent just useless deviations. It'd be interesting to see the source of this nice question as well. Remark. The same idea for evaluating such sums is used in the elementary proof of the fact that $ \zeta(2) = \pi^2/6$.