What are the allowed wavenumbers in the finite size system? Usually, we introduce wavenumber $\textbf{q}$ by Fourier transform, for example, an operator $A_{\textbf{q}}=1/\sqrt{N}*\sum_{i}e^{i \textbf{q}\cdot \textbf{r}_{i}}A_{i}$, where $N$ is number of sites, $i$ denotes the site index. For a lattice or finite system with regular discrete $i$ (e.g. finite (spin) chain), $\textbf{r}_i$ takes integer number, then $A_\textbf{q}=A_{\textbf{q}+2\pi\hat{1}}$. So $q$ always takes value from $(0,2\pi)$.
From solid state physics textbook, $i$ component of $\textbf{q}$ takes quantized values as $2\pi n/N_i$ once we used period boundary condition, here $i$ denotes $x, y, z$. With period boundary condition, we have translation symmetry and $\textbf{q}$ is proportional to momentum.
But what if in finite size system, what are the allowed values of $q$?
For example:
1) a finite chain, when the length $L$ takes finite value, such as 50. 
2) a two-leg ladder, suppose there is translation symmetry along the
leg direction $x$, then we have $q_x=2\pi n/N_x$, but does it make
sense to define $q_y$? (On the other hand, the Brillouin zone is one
dimension defined by period along $x$ direction.)
This ladder case can also be extended to bilayer case, such as bilayer square lattice. Then does it make sense to define $q_z$ along inter-layer direction?
If this $q_y$ in ladder can be defined, how to understand it? It is not the momentum anymore.
 A: You already have your answer in the question. You said $0<q<2\pi$. Plus $q=n\frac{2\pi}{N}$ (here you assumed a distance 1 between your spins otherwise it would be $n\frac{2\pi}{Na}=n\frac{2\pi}{L}$ where $a$ is the distance between spins). Since n is a positive integer, $q<2\pi$ induces $n\leq N-1$.
edit :
Thanks to your comment I think I get what you ask. 
The problem is simply that to work with fourier transform you need a function defined on an infinite number of site. Otherwise, you won't get the inverse fourier transform. I'll stick to a continuous function $f(x)$ and its fourier transform $F(q)$ here for clarity.
$\begin{align} F(q) &= \int\limits_{-\infty}^{+\infty}dx \, f(x) e^{-iqx} \\ &= \int\limits_{-\infty}^{+\infty}dx \, e^{-iqx} \int \frac{dk}{2\pi}F(k) e^{+ikx} \\ &= \int \frac{dk}{2\pi}F(k) 2\pi \delta(q-k)=F(q) \end{align}$
I needed to use $\int\limits_{-\infty}^{+\infty}dx \, e^{ix(k-q)}=2\pi\delta(q-k)$. But if you want to do so on a finite domain you'll get $\int\limits_{-L/2}^{+L/2}dx \, e^{ix(k-q)}=L\text{sinc}(\frac{qL}{2})$ which gives back the delta function you need only for $L\rightarrow{} \infty$. Saying that your function is periodic is a way to define it on an infinite domain. You could also say that your function is $0$ everywhere outside the domain of interest.
A: In a coordinate where you have open boundary conditions, the problem is equivalent to a more familiar one: that of a particle in a box. Which wave numbers are allowed there? Well, it's precisely the $k=\frac{n\pi}{L}$ mentioned in @E. Bellec's answer. Multiply by $\hbar$ and you have a physical momentum. It might not have all the properties of the crystal momentum along another coordinate with periodic boundary conditions, but it certainly is a momentum.
Now, taking your ladder as an example - yes, it makes sense to define $k_y$. Similarly for the out-of-plane component $k_z$ in the bilayer example. However, for these cases, the gap between the two associated energy levels is usually quite large (since L is small), and the higher level is often considered to be "frozen out". Hence, a spin ladder can often be treated as a quasi-1D problem, and a bilayer as a quasi-2D problem.
