An $n$-tuple of nonnegative integers $(x_{1}, \ldots , x_{n})$, with no restrictions on their common denominator, will be called good if there exists a sequence of moves which result in $n-1$ zeros left on the blackboard. We would like to find the set of all good $n$-tuples. I claim that the good $n$-tuples $(x_{1}, \ldots , x_{n})$ are those for which $\frac{x_{1}}{g}+\ldots+\frac{x_{n}}{g}$ is a power of $2$, where $g=GCD(x_{1}, \ldots , x_{n})$. We first state a few observations. The sum $x_{1}+\ldots+x_{n}$ is invariant over moves. $(x_{1}, \ldots , x_{n})$ is good if and only if $(Nx_{1}, \ldots , Nx_{n})$ is good for any positive integer $N$. An odd integer $m$ divides $x_{1}-x_{2},2x_{2},x_{3},\ldots,x_{n}$ if and only if $m$ divides $x_{1},x_{2},x_{3},\ldots,x_{n}$. This implies in particular that the gcd of the numbers on the board either stay the same or differ by factors of $2$ between moves. Suppose that $(x_{1}, \ldots , x_{n})$ is good. Let $m$ be the greatest odd integer which divides $x_{1}+\ldots+x_{n}$. The 3rd observation above implies that $m$ divides each of $x_{1},\ldots,x_{n}$. Hence $m$ divides $g$ and $\frac{x_{1}}{g}+\ldots+\frac{x_{n}}{g}$ is a power of $2$. Suppose that $\frac{x_{1}}{g}+\ldots+\frac{x_{n}}{g}$ is a power of $2$; we prove that $(x_{1}, \ldots , x_{n})$ is good. By the 2nd observation above, it suffices to prove that $(\frac{x_{1}}{g},\ldots,\frac{x_{n}}{g})$ is good, so let us assume $g=1$. If $x_{1}+\ldots+x_{n}=1$, then WLOG $x_{1}=1$ and the rest are zero, and there is nothing to prove. Let $x_{1}+\ldots+x_{n}>1$. Since $x_{1}+\ldots+x_{n}$ is even, we can perform a series of moves so that no odd number is left on the board (performing a move on two odd numbers gives two even numbers). Suppose that the even numbers left on the board now are $x_{1}',\ldots,x_{n}'$; since $\frac{x_{1}'}{2}+\ldots+\frac{x_{n}'}{2}$ is a power of $2$, the induction hypothesis tells us that $(\frac{x_{1}'}{2}, \ldots , \frac{x_{n}'}{2})$ is good; and $(x_{1}, \ldots , x_{n})$ is good by the 2nd observation.

lets observer the operation from the final move. at the last move we must have $(x,x,0,0,0,...)$ and if this is the $n$th move we must have such construction at the $n-1$th move $(x,y,y,0,0,..)$ where $2y=x$ which implies that $x$ is even or there are just two numbers on the board which have to be $(1,1)$ every time we move back one step it is easy to observe that if $p$ is an odd prime divisor of $x$ then $p$ is also a prime divisor of $\frac{x}{2^k}$ so such $p$ does not exist therefore the sum of all the numbers on the board has to be a power of 2. now using induction on the power of $S$ which $S$ is the sum of all the numbers on the board can easily solve the question EDIT: wrong induction fixed now