I can't tell if the question means "$2\times 3$ in only that orientation", or it means "$2\times 3$ or $3\times 2$". I've assumed the latter in what follows. If the sheet contains the "slash" R X X R then by repeated application of the rule, you must have an infinite square tiling tiling of the $3\times 3$ pattern R X X X R X X X R Similarly, the red tiles could run along northeasterly diagonals instead. In either case, it should be obvious that any $9\times 12$ square will contain $36$ red tiles. Any $9\times 11$ sub-square loses a strip of size $9\times 1$, containing $3$ red tiles, from the left or right of the $9\times 12$ area, leaving ${\bf 33}$ red tiles in total. Now we need to show that no other tiling of the sheet is possible. If the sheet contains the "dash" R R X X but no "slash", then it contains ? X X ? X X X X X R R X X X X X ? X X ? but then there's only at most one red tile in the top $2\times 3$ rectangle -- FAIL. Finally, if the sheet contains no "slash" or "dash" (or their $90^\circ$ rotations) then any red is surrounded by eight non-red tiles, which clearly fails the rule given in the question.

This problem was also discussed here: https://artofproblemsolving.com/community/c6h2253263p17375869

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