, 2025

Yes, this is possible. Consider the spiral depicted below and write consecutive integers along the spiral:

We claim that this works. Consider any two cells $A$ and $B$ sharing exactly one vertex. Consider the $2 \times 2$ square containing $A$ and $B$. If the $2 \times 2$ square contains a "corner" of the spiral then for some $n$ and $k$ the numbers in that $2 \times 2$ square are arranged in the following way (up to rotation or reflection):
$$\begin{array}{|c|c|} \hline n+1 & k \\ \hline n & n-1 \\ \hline \end{array}$$
and therefore the conditions are satisfied no matter which opposite cells of the $2 \times 2$ square $A$ and $B$ are. Indeed, if $A$ and $B$ contain $n-1$ and $n+1$, then the good cell is the one containing $n$. If $A$ and $B$ contain $n$ and $k$ and $n < k$ then $n < n+1 < k$ and the good cell is the one containing $n+1$. If $A$ and $B$ contain $n$ and $k$ and $n > k$ then $k < n-1 < n$ and the good cell is the one containing $n-1$.

Otherwise, the numbers are arranged in the following way (again, up to rotation or reflection):
$$\begin{array}{|c|c|} \hline k+1 & k \\ \hline n & n+1 \\ \hline \end{array}$$

for some $n, k$, and again, the conditions are satisfied. Indeed, without loss of generality, assume $n < k$. Then $n < n+1 < k < k+1$. If $A$ and $B$ contain $n$ and $k$ then the good cell is the one containing $n+1$. Otherwise, $A$ and $B$ contain $n+1$ and $k+1$, and the good cell is the one containing $k$.

Alternatively, one can notice that condition (iii) from the problem statement means that we can orient each square $2 \times 2$ according to the increasing numbers as suggested in the picture below:

With this observation, it's relatively easy to check that the spiral construction satisfies this.