Solution:

If a prime $p$ divides $a + b$ and $a - b$, then it divides $2a$ and $2b$, so if $p$ is odd, it divides $a$ and $b$. Thus if an odd prime $p$ divides $k$, then it must divide all the original numbers including $1$. So $k$ must be a power of $2$. Note that $k, k \rightarrow 0, 2k \rightarrow 2k, 2k$ and $k, k, k \rightarrow 0, k, 2k \rightarrow k, k, 2k \rightarrow 0, 2k, 2k \rightarrow 2k, 2k, 2k$. So (by a trivial induction) if we get all the numbers equal to $k$, then we can get them all to equal $2k$. Finally, note that we can never decrease the largest number on the board, so the answer must be all powers of $2$ greater than some minimum, which must be at least $n$.

We use induction to show that if $2^{m}$ is the smallest power of $2$ which is $\geq n$, then we can get all numbers equal to $2^{m}$. Note that $0, k \rightarrow k, k \rightarrow 0, 2k$, so with a zero we can double each member of any set of numbers as often as we wish and finally convert the zero. For example, we could convert $0, 2, 4$ to $8, 8, 8$. It is convenient to take the induction hypothesis as $S_{n}$: we can convert $1, 2, \ldots, n$ to $0, 2^{k}, 2^{k}, \ldots, 2^{k}$, where $2^{k}$ is the smallest power of $2$ which is $\geq n$.

We show first that $S_{n}$ is true for $n \leq 8$. For $n = 3$, we take $1, 3 \rightarrow 2, 4$, then $2, 2 \rightarrow 0, 4$. For $n = 4$, we ignore the $4$ and use the case $n = 3$. For $n = 5$, we take $3, 5 \rightarrow 2, 8$. Then $2, 2 \rightarrow 0, 4$. Then we use the $0$ to convert the remaining powers of $2$ ($1, 4, 4$) to $8$. For $n = 6$, we take $2, 6 \rightarrow 4, 8$ and $3, 5 \rightarrow 2, 8$, then $4, 4 \rightarrow 0, 8$. Finally, we use the $0$ to convert $1$ and $2$ to $8$. For $n = 7$, we take $1, 7 \rightarrow 6, 8$, then $2, 6 \rightarrow 4, 8$, then $3, 5 \rightarrow 2, 8$, then $4, 4 \rightarrow 0, 8$, then $2, 6 \rightarrow 4, 8$ and finally use the $0$ to convert the remaining $4$ to $8$.

Let $n = 2^{a} + b$, where $0 < b \leq 2^{a}$ and assume $S_{m}$ is true for all $m < n$. If $b = 1$, we convert the pair $2^{a} - 1$, $2^{a} + 1$ to $2, 2^{a + 1}$. We have $2^{a} - 2 > 2$, so by induction we can convert $1, 2, \ldots, 2^{a} - 2$ to $0, 2^{a}, \ldots, 2^{a}$. Now all the numbers except $0$ are powers of $2$ and we can use the $0$ to convert them each to $2^{a + 1}$. Similarly, if $b = 2$, we convert $2^{a} - 1$, $2^{a} + 1$ to $2, 2^{a + 1}$ and $2^{a} - 2$, $2^{a} + 2$ to $4, 2^{a + 1}$ and then proceed as in the previous case. If $3 \leq b < 2^{a}$, then we start by converting the pairs $(2^{a} + b, 2^{a} - b)$, $(2^{a} + b - 1, 2^{a} - b + 1)$, $(2^{a} + b - 2, 2^{a} - b + 2)$, $\ldots$, $(2^{a} + 1, 2^{a} - 1)$. That gives some $2^{a + 1}$s and $2, 4, \ldots, 2b$. Now by $S_{b}$ we can convert $2, 4, \ldots, 2b$ to $0, 2^{a + 1}, \ldots, 2^{a + 1}$. The remaining numbers $1, 2, \ldots, 2^{a} - b - 1$ can either be converted to powers of $2$ by $S_{2^{a - b - 1}}$ (if $2^{a} - b - 1 \geq 3$) or are already powers of $2$. Finally we use the $0$ to bring all powers of $2$ up to $2^{a + 1}$. In the case $b = 2^{a}$, we ignore $2^{a} + b (= 2^{a + 1})$ and use the case $b - 1$ to convert the others.