APMO

Let $n$ be a positive integer relatively prime to $2$ and $3$. We may study the whole process modulo $n$ by replacing divisions by $2,3,4$ with multiplications by the corresponding inverses modulo $n$. If at some point the original process makes all the numbers equal, then the process modulo $n$ will also have all the numbers equal. Our aim is to choose $n$ and an initial configuration modulo $n$ for which no process modulo $n$ reaches a board with all numbers equal modulo $n$. We split this goal into two lemmas.

Lemma 1. There is a $2\times 3$ board that stays constant modulo $5$ and whose entries are not all equal.

Proof. Here is one such a board:

3	1	3
0	2	0

The fact that the board remains constant regardless of the choice of squares can be checked square by square. $\square$

Lemma 2. If there is an $r\times s$ board with $r\ge 2,s\ge 2$, that stays constant modulo $5$, then there is also a $kr\times ls$ board with the same property.

Proof. We prove by a case by case analysis that repeatedly reflecting the $r\times s$ with respect to an edge preserves the property:

- If a cell had $4$ neighbors, after reflections it still has the same neighbors. - If a cell with $a$ had $3$ neighbors $b,c,d$, we have by hypothesis that $a\equiv 3^{-1}(b+c+d)\equiv 2(b+c+d)(\mathrm{mod}5)$. A reflection may add $a$ as a neighbor of the cell and now $$4^{-1}(a+b+c+d)\equiv 4(a+b+c+d)\equiv 4a+2a\equiv a(\mathrm{mod}5)$$ - If a cell with $a$ had $2$ neighbors $b,c$, we have by hypothesis that $a\equiv 2^{-1}(b+c)\equiv 3(b+c)(\mathrm{mod}5)$. If the reflections add one $a$ as neighbor, now $$3^{-1}(a+b+c)\equiv 2(3(b+c)+b+c)\equiv 8(b+c)\equiv 3(b+c)\equiv a(\mathrm{mod}5)$$ - If a cell with $a$ had $2$ neighbors $b,c$, we have by hypothesis that $a\equiv 2^{-1}(b+c)(\mathrm{mod}5)$. If the reflections add two $a$'s as neighbors, now $$4^{-1}(2a+b+c)\equiv (2^{-1}a+2^{-1}a)\equiv a(\mathrm{mod}5)$$

In the three cases, any cell is still preserved modulo $5$ after an operation. Hence we can fill in the $kr\times ls$ board by $k\times l$ copies by reflection. $\square$

Since $2\mid 2018$ and $3\mid 2019$, we can get through reflections the following board:



By the lemmas above, the board is invariant modulo $5$, so the answer is no.