Tessellate S.T.E.M.S. (2019) - Mathematics - Category A (School) - Set 2 - Objective Problem 4

First, we realize that the greatest theoretically possible sum is $2 \cdot 16 = 32$ . But if there are only 2s, then the sum of all entries would be divible by 4. If we put a 1 somewhere, the total sum won't be divisible by 4, but all $3 \times 3$ sub-squares that contain this 1 will have a sum of $2 \cdot 8 + 1 \cdot 1 = 17$ , which is not divisible by 4. If we now take another 1 and put it and the other 1 into the central $2 \times 2$ sub-square both 1s will be in all $3 \times 3$ sub-squares, so each of these sums will be $2 \cdot 7 + 1 \cdot 2 = 16$ which is divisible by 4. The total sum will be $2 \cdot 14 + 1 \cdot 2 = \boxed{30}$ and will not be divisible by 4.