Euler characteristic of a manifold is odd

$\newcommand{\Q}{\mathbb{Q}}$Poincaré duality tells you that there are non-degenerate pairings $H^i(M) \otimes H^{n-i}(M, \partial M) \to \Q$ for all $0 \le i \le n$.

Using the long exact sequence of the pair $(M, \partial M)$, the known facts that $H^n(M) = 0$ and $H^n(M,\partial M) = \Q$ (Poincaré duality again), and the description of $H^*(\partial M) = H^*(S^{n-1})$, one can deduce that $H^i(M) \cong H^i(M, \partial M)$ for all $i \le n-1$. The non-degenerate pairing mentioned above thus tells you that $\dim H^i(M) = \dim H^{n-i}(M)$ for all $0 < i < n$.

Write $n = 4k+2$ for convenience. We can therefore start to compute the Euler characteristic, pairing terms appropriately: \begin{align} \chi(M) & = \sum_{i=0}^n (-1)^i \dim H^i(M) \\ & = \dim H^0(M) + \sum_{i=1}^{2k} (-1)^i \dim H^i(M) - \dim H^{2k+1}(M) \\ & \qquad+ \sum_{i=2k+2}^{4k+1} (-1)^i \dim H^i(M) + \dim H^{4k+2}(M) \\ & = 1 + 2 \sum_{i = 1}^{2k} (-1)^i \dim H^i(M) - \dim H^{2k+1}(M). \end{align}

So we just need to show that $\dim H^{2k+1}(M)$ is even. Because it's the middle dimension, we have a non-degenerate pairing $H^{2k+1}(M) \otimes H^{2k+1}(M, \partial M) \to \Q$. Composing with the isomorphism $H^{2k+1}(M) \cong H^{2k+1}(M,\partial M)$ we get a non-degenerate pairing $H^{2k+1}(M) \otimes H^{2k+1}(M) \to \Q$, given by $\alpha \otimes \beta \mapsto \langle \alpha \cup \beta, [M] \rangle$.

Since $2k+1$ is odd, this pairing is skew-symmetric (and non-degenerate). In particular, $H^{2k+1}(M)$ is a symplectic vector space, and therefore it is necessarily even-dimensional. Plugging this in the formula for $\chi(M)$ above, we finally get that $\chi(M)$ is odd.