Solution:

There cannot be two red squares with a common side. For consider as the diagram shows we can immediately conclude that the squares with a $\ast$ are not red, but now the bold rectangle has at most 1 red square. Contradiction.

Consider a red square. One of the two diagonally adjacent squares marked $\ast$ must be red. But it is now easy to show that all red squares on that diagonal are red and that the other red squares are those on every third parallel diagonal line. Any $9 \times 11$ rectangle must have just three such diagonals on a 9 cell border row, and hence just 3 red cells in that border row. But the remaining $9 \times 10$ rectangle can easily be partitioned into fifteen $3 \times 2$ rectangles, each with 2 red squares.