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 $r \geq 2, s \geq 2$, that stays constant modulo $5$, then there is also a $k r \times l s$ 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 $a \equiv 3^{-1}(b+c+d) \equiv 2(b+c+d)\ (\bmod 5)$. A reflection may add $a$ as a neighbor of the cell and now
$$4^{-1}(a+b+c+d) \equiv 4(a+b+c+d) \equiv 4a+2a \equiv a\ (\bmod 5)$$
- If a cell with $a$ had $2$ neighbors $b, c$, we have by hypothesis that $a \equiv 2^{-1}(b+c) \equiv 3(b+c)\ (\bmod 5)$. If the reflections add one $a$ as neighbor, now
$$3^{-1}(a+b+c) \equiv 2(3(b+c)+b+c) \equiv 8(b+c) \equiv 3(b+c) \equiv a\ (\bmod 5)$$
- If a cell with $a$ had $2$ neighbors $b, c$, we have by hypothesis that $a \equiv 2^{-1}(b+c)\ (\bmod 5)$. If the reflections add two $a$'s as neighbors, now
$$4^{-1}(2a+b+c) \equiv (2^{-1}a+2^{-1}a) \equiv a\ \quad(\bmod 5)$$

In the three cases, any cell is still preserved modulo $5$ after an operation. Hence we can fill in the $k r \times l s$ 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.