The answer is $\boxed{\text{every even }n\text{ and }1}$. Solution. If $n=1$ it's trivial - no moves needed. If $n\geq 3$ is odd then pick some $n$ distinct reals in $(2,3)$ as our starting point. Now it's easily seen that except at the initial state, the maximal value on the board must appear an even number of times, therefore we cannot end up with all numbers equal If $n$ is even then we'll show that this is always possible by induction on $n$. The base case $n=2$ is trivial. For the inductive step, suppose that we've proved the case $n=k$. Now suppose we have an $k+2$-tuple $(a_1,a_2,\ldots,a_k,a_{k+1},a_{k+2})$. Use the inductive hypothesis on $a_1,\ldots,a_k$ to make them all equal, then apply a move to $a_{k+1},a_{k+2}$, so that our tuple is now of the form $(x,x,\ldots,x,y,y)$. Now do the following moves: (the number over the arrow is the two positions chosen) $$(x,x,x,x,\ldots,x,y,y) \xrightarrow{(3,k+1)} (x,x,xy,x,\ldots,x,xy,y)$$$$\xrightarrow{(4,k+1)} (x,x,xy,x^2y,\ldots,x,x^2y,y)$$$$\xrightarrow{(5,k+1)} \cdots \xrightarrow{(k,k+1)} (x,x,xy,x^2y,\ldots,x^{k-2}y,x^{k-2}y,y)$$$$\xrightarrow{(k+1,k+2)} (x,x,xy,x^2y,\ldots,x^{k-2}y,x^{k-2}y^2,x^{k-2}y^2)$$$$\xrightarrow{(i,(k+3)-i)\text{ for every i}} (x^{k-1}y^2,x^{k-1}y^2,x^{k-1}y^2,x^{k-1}y^2,\ldots,x^{k-1}y^2,x^{k-1}y^2)$$therefore we've proved the case $n=k+2$ as well. $\blacksquare$