Physical interpretation of the Gibbs-Duhem equation
If you write down the equation in terms of the very definition of $\mu_i=\frac{\rm dG}{{\rm d}n_i}$ and impose ${\rm d}n_\text{totoal}=0$, it's just the conservation of the energy potential ${\rm d}G=0$.



Say you own a theme park where, every hour, each adult has to pay ${\rm d}\mu_a=\$7$ (money=energy) and each kid has to pay ${\rm d}\mu_k=\$3$. 
We assign positive signs to when they put money out of their pockets. 

If there are $n_a=10$ adults and $n_k=5$ kids in the park, and if this number of people stays fixed (${\rm d}n_\text{totoal}=0$ over some time), then they together have to spend 

$$n_a\cdot {\rm d}\mu_a + n_k\cdot {\rm d}\mu_k = \$70 + \$15 = \$85$$ 

Now say you, the park owner, are not allowed to actually make any money (${\rm d}G=0$) and you relax the condition that the kids have to pay any money at all. That is, now just the adult have to pay ... and since the money has to go somewhere, the kids are on the receiving end. Then the ten parents still spend $\$70$ per hour and now the five kids will split that money among each other. Each kids receives
$$-{\rm d}\mu_k = \frac{1}{n_k}\cdot n_a\cdot {\rm d}\mu_a = \frac{1}{5}\cdot \$70 = \$14$$ 



The comparison with "money per time" lacks in that here people don't bring in money just from coming to the park (as it's the case with particles coming into the system and energy). But I hope this clear up the meaning of factor $\frac{1}{n_k}$ clear.