Is every rational realized as the Euler characteristic of some manifold or orbifold?

Products of 2-orbifolds with manifolds will do the trick. There are 2-orbifolds of Euler characteristic $1/n$ (take a quotient of $S^2$ by a rotation of order $2n$). Then take a product with a manifold of Euler characteristic $m \in \mathbb{Z}$ to get all rationals.

The answer for connected 2-dimensional orbifolds is no. Euler characteristic is $$\chi(O)=\chi(M)-\sum\left(1-\frac{1}{q}\right)-\frac{1}{2}\sum\left(1-\frac{1}{p}\right),$$ where $p,q\geq 2$ are integers, and $\chi(M)$ is the characteristic of the surface of the orbifold. It immediately follows that $\chi(O)\leq 2$ for all 2D orbiforlds, and as $1-1/q\geq 1/2$, $1-1/p\geq 1/2$, most rational numbers will never occur (on any closed interval which does not contain half-integers, there are only finitely many of these numbers).