In ordinary substances, we have the system in thermal equilibrium with the environment, and with a fixed number of particles. However we go to the grand canonical system for efficiency, where we do not have to restrict ourselves to a fixed particle number $N_{0}$, and that simplifies calculations enormously. In this, the system is coupled to the environment which is kept at a fixed Temperature $T$ and chemical potential $\mu$.
In reality, in our actual problem,we are in a "canonical" system, and the quantities $T$ and $N_{0}$ are fixed. However we go to the grand canonical system, and work with a chemical potential $\mu$, along with the fixed $T$, that makes sure to produce an $\textbf{average particle number} \ <N>$ equal to $N_{0}$, in the grand canonical system. Thus, the chemical potential is brought forth in an artificial way.
In a way, we have two degrees of freedom, actually $T$ and $N_{0}$, but we chose to work with independent parameters $T$ and $\mu$ while getting our results from the grand canonical theory.
Now, in the grand canonical ensemble, each state characterized by a particular total energy $E$ and total particle number $N$, has the probability of appearing
\begin{equation}
\rho\sim e^{-\beta E+\beta \mu N}
\end{equation}
Now, let's deal with fermions first: as $T\rightarrow 0$, the probability of having higher energy states becomes negligible. However, with $\mu>0$ as $T\rightarrow 0$, we the probabilities become exponentially larger as we keep adding more particles. Combining the two facts, we predominantly end up with states with simultaneously "low" total energy and with "high" number of particles. (The use of the terms "high" and "low" is subjective, but should be clear from context.). But this does not give a divergingly large contribution for fermions, because with a particular "small" total energy, due to the Pauli Exclusion Principle, the total number of particles cannot exceed a certain value. Hence as $T\rightarrow 0$, we're required to have a positive value of $\mu$. There are only so many states which give a meaningful contribution.[See note in the end.]
If we end up averaging over the states that do contribute, we'll find that they produce a given average number of particles $N_{0}$, with a suitable choice of $\mu(T=0)$. (That is of course, exactly what is done).
Now as the temperature $T$ increases, higher energy states start becoming more probable, and if $\mu$ were to stay constant or increase, we would start having a problem, as there would be more states with more energy and more number of particles that give a meaningful contribution, and thus the average number of particles $<N>$ in the system would shoot up. But of course, the whole point is to keep the particle number $<N>=N_{0}$,a constant. Thus we have to postulate that $\mu$ decreases and becomes negative,(very rapidly, as we'll argue in a moment).
When $T$ is very large and thus $\beta$ is very small , we argue that $\mu$ is negative , which we write as $\mu=-|\mu|$, and thus, now,
\begin{equation}
\rho\sim e^{-\beta E-\beta |\mu| N}
\end{equation}
As $\beta\rightarrow 0$, higher energy states are more probable, but we need to ensure that $|\mu|$ increases at a tremendously fast rate so that $-\beta|\mu|$ is a very large negative quantity(even when $\beta\rightarrow 0$) so that states with very high number of particles give a very negligible contribution. Thus only states with a broad spectrum of energies and simultaneously with low particle number contribute. For this to be possible,again, this would give on average , the same value for the average number of particles $<N>=N_{0}$ in the system( see the Note in the end).
Thus , for all temperatures, to keep the particle number constant, essentially, we end up varying $\mu$ suitably.
The high temperature limit argument is the same for bosons as well, but as pointed out in Lubos Motl's answer, when $T\rightarrow 0$, we absolutely cannot have a positive value of $\mu$ because the Pauli Exclusion Principle does not hold for bosons, and we would get a violent divergence for the average $<N>$ as $T\rightarrow 0$. Thus, for bosons, we need to start with a value of $\beta\mu=0$ as $\beta\rightarrow \infty$.
[In fact, there is divergence nonetheless for $T\leq T_{critical}$ and stricty speaking our formalism holds only for temperature values above the (very small)critical value. $\beta\mu\approx 0$ at $T=T_{critical}$, and formally, it is also taken $0$ for all values of $T$ below the critical value. For example, as shown in Pathria Figure 7.2.]
Thus overall, in case of both fermions and bosons, we can use the average particle number $<N>$ that we get, as the value of the actual fixed value of $N_{0}$ in the original problem.
[Note: Here , we basically are using a density of states argument for states simultaneously at a particular total energy $E$ and particular total $N$ value. After we get the probability of the state appearing from the standard probability of states formula (which is the result of entropy maximisation, in a way), we need to find out how many such states are actually there, from the density of states, as a function simultaneously of $E$ and $N$. That can in principle be estimated from the individual particle density of states, essentially a combinatorial argument].
