The short answer is that you have to work with particle densities, namely, $$n=N/V,$$ where $N$ is the number of particles, and $V$ is the volume of your system.
The long answer is as follows.
Working in the thermodynamic ensemble with a fixed chemical potential $\mu$, and knowing that the Bose-Einstein distribution gives the average number density of non-condensed particles $$n_{nc} = \frac{1}{V} \sum_{\boldsymbol k\neq 0}\frac{1}{\exp(\beta [\hbar^2 \boldsymbol k^2/2m - \mu])-1},$$
where $\beta = 1/k_B T$ is the inverse temperature, $k_B$ is the Boltzmann coefficient, $m$ is the mass of the particles, and $\hbar\boldsymbol k$ is the momentum. Fixing the temperature $T$ and the number of excited particles $n_{nc}$ allows one to find the chemical potential for the situation at hand.
Note that $\mu = - \infty$ gives a very small particle density, which increases with increasing $\mu$.
However, requiring a low enough temperature in combination with a high enough density, gives a problem. Namely, satisfying the equation means that the chemical potential has to be positive. This means that some low-momentum mode is occupied negatively, as $$\frac{1}{\exp(-\beta\mu)-1}\simeq \frac{1}{-\beta\mu},$$ for small $\beta\mu$. We thus conclude that positive $\mu$'s are not allowed.
This paradox is resolved by realizing that the correct formula for the total number of particles is actually $$n = n_0 + n_{nc}.$$ Therefore, if after fixing the total number of particles $n$ and the temperature $T$ one sees that $n_{nc}$ with $\mu=0$ is still less than $n$, one has to conclude that the rest of the particles reside in the condensate, and hence $n_0\neq 0$.
All this is standard textbook knowledge. It is elaborated on, for example, in the book Bose–Einstein Condensation in Dilute Gases by Pethick and Smith.
