In a box with sides of length $L$, the energy eigenvalues depend on the boundary conditions. For periodic boundary conditions, they are $$E_{n_x,n_y,n_z}=\frac{\hbar^2}{2m}\left(\frac{2\pi}{L}\right)^2(n_x^2+n_y^2+n_z^2)\tag{1}\label{1}$$ where $(n_x,n_y,n_z)$ are integers, but for reflecting boundary conditions, they are $$E_{n_x,n_y,n_z}=\frac{\hbar^2}{2m}\left(\frac{\pi}{L}\right)^2(n_x^2+n_y^2+n_z^2)\tag{2}\label{2}$$ where $(n_x,n_y,n_z)$ are positive integers. In most cases, this makes no difference, as they both yield the same density of states. However, in a Bose-Einstein condensate, where a macroscopic number of particles are in the ground state, it seems to have an effect.
For example, in $(\ref{1})$, the ground state is $E_{0,0,0}=0$, whereas in $(\ref{2})$ it is $E_{1,1,1}=\frac{3\hbar^2\pi^2}{2mL^2}$. This is just an arbitrary energy shift. Where a difference arises is the energy gap to the first excited state. For $(\ref{1})$, this is $E_{1,0,0}-E_{0,0,0}=\frac{2\hbar^2\pi^2}{mL^2}$, but for $(\ref{2})$ it is $E_{2,1,1}-E_{1,1,1}=\frac{3\hbar^2\pi^2}{2mL^2}$. According to Bose-Einstein statistics, the two otherwise identical gases should therefore have a different proportion of particles in the first excited state. How is this reconciled?
Edit: to derive $(\ref{1})$, you assume wave function (for a box centred on the origin) $\psi(\boldsymbol{r})=\frac{1}{\sqrt{L^3}}e^{i\boldsymbol{k}\cdot\boldsymbol{r}}$ and impose periodic boundary conditions, e.g. $\psi(L/2,y,z)=\psi(-L/2,y,z)$. To derive $(\ref{2})$, you use 3D square well wavefunction (now with a corner at the origin) $\psi(\boldsymbol{r})=\left(\frac{2}{L}\right)^{3/2}\sin(k_xx)\sin(k_yy)\sin(k_zz)$ and impose that it must be zero at the walls.