For each $ n$, $ g\left(n\right)$ equals to $ 2$ ,when $ n$ is a power of $ 2$, and $ 3$ when otherwise. First, we will do the easier side, that is, we have the steps of operations that leave $ g\left(n\right)$ numbers on the board. Suppose that $ n$ is a power of $ 2$. We can just fix $ 0$ and $ n$ so we can erase $ 1, 3, ..., n-1$. Then, like induction, we can erase $ 2\cdot 1,...,2\cdot (n-2)/2$ and so on. (We erase those of $ 1(mod 2)$ then $ 2(mod 4)$ then ... then $ 2^{m-1} (mod 2^{m})$) By this way, we can erase all but $ 0$ and $ n$ which is what we wanted. When in the case that $ n$ is not a power of $ 2$, let $ n=2^{m}+a$ where $ a, m$ is a natural number with $ 0 < a < 2^{m}$. We will not erase $ 0,2^{m}$ nor $ n$. We then erase $ n-1, n-2, ... 2^{m}+1$ respectively (or not erase any of them if $ a=1$) and erase $ 1, ... , 2^{m}-1$ in the same way as the first case. By this way, we can leave $ 0, 2^{m}$ and $ n$ as we claimed. Now, for the latter side, that is, we will show that $ g\left(n\right)$ can't be less than we claimed. Notice that $ 0$ and $ n$ can't be erase -- this yields $ g\left(n\right) \geq 2$ The last part to be tackled is when $ n$ is not a power of $ 2$. Since $ n$ is not a power of $ 2$, we have a prime $ p>2$ that $ p\mid n$ Suppose, for sake of contradiction, that $ g\left(n\right)=2$. Then, the left numbers are $ 0$ and $ n$. If there is a series of steps that can leave only $ 0$ and $ n$, then we can do it backward by writing the integer number that is arithmetic mean of two different numbers on the board (which, initially, are $ 0$ and $ n$) and repeat this process until we have all the numbers $ 0,1,...,n$ But notice that, by writing in this fashion, all the new number we get must still be divisible by $ p$ In particular, we can't get $ 1$ => contradiction!