There are $(n-1)^2$ different $2\times 2$ squares and sum of their numbers ranges from $0$ to $4n$. So we can connect at most $4n+1$ values to $(n-1)^2$ different squares, thus $4n+1\ge(n-1)^2\implies n\in\lbrace 3,4,5,6\rbrace$ $n=6$ 0,0,0,0,0,1 1,0,0,2,1,1 4,0,6,1,3,3 0,6,0,4,5,3 6,6,5,6,4,4 6,6,6,5,6,6 $n=5$ 0,0,0,2,0 0,1,1,0,2 0,4,5,4,1 0,4,3,5,5 0,2,5,5,5 $n=4$ 0,1,2,3 0,0,3,4 2,3,4,4 4,4,1,2 $n=3$ 0,0,0 1,0,2 2,3,0

WolfusA wrote: There are $(n-1)^2$ different $2\times 2$ squares and sum of their numbers ranges from $0$ to $4n$. So we can connect at most $4n+1$ values to $(n-1)^2$ different squares, thus $4n+1\ge(n-1)^2\implies n\in\lbrace 3,4,5,6\rbrace$ $n=6$ 0,0,0,0,0,1 1,0,0,2,1,1 4,0,6,1,3,3 0,6,0,4,5,3 6,6,5,6,4,4 6,6,6,5,6,6 $n=5$ 0,0,0,2,0 0,1,1,0,2 0,4,5,4,1 0,4,3,5,5 0,2,5,5,5 $n=4$ 0,1,2,3 0,0,3,4 2,3,4,4 4,4,1,2 $n=3$ 0,0,0 1,0,2 2,3,0 I think your example for "the cases n=4 and n=6" is mistaken. Do you have any other examples? Edit: I suppose these works: For n=6; 5,5,6,6,6,6 5,5,5,5,6,6 3,2,4,3,4,3 5,4,3,2,2,1 0,0,1,1,1,1 0,0,0,0,1,1 For n=4; 0,0,0,0 0,0,1,1 1,2,2,2 2,2,3,3

Just solve 4n+1>=(n-1)^2