We label the rows from 1 to $m$ and the columns from 1 to $n$.
If $m$ and $n$ are not congruent to 2 modulo 3, i.e. $m = 3a + r_1$ and $n = 3b + r_2$ with $r_1, r_2 \in \{0, 1\}$, we can tile the rectangle formed by the first $3a$ lines and $3b$ columns with $a \cdot b$ disjoint $3 \times 3$ squares. Each of these squares needs to contain at least 2 red squares, therefore one needs to paint red at least $2ab = 2 \cdot \lfloor \frac{m}{3} \rfloor \cdot \lfloor \frac{n}{3} \rfloor$ squares. On the other hand, this number is sufficient: paint red those unit squares whose row number is a multiple of 3 (i.e. 3, 6, ..., $3a$) and whose column number belongs to the set $\{2, 3, 5, 6, ..., 3b-1, 3b\}$. It is easy to check that each $3 \times 3$ square contains exactly two red squares.

The same answer (with the exact same arguments) remains valid if $m \equiv 2 \pmod{3}$ and $n \equiv 0, 1 \pmod{3}$ (the answer remains the same in the case $n \equiv 2 \pmod{3}$ and $m \equiv 0, 1 \pmod{3}$).

Let us now tackle the case $m = 3a + 2$, $n = 3b + 2$. We prove that in this case the answer is $2ab + \min\{a, b\}$.

Let us call a zone a $5 \times 5$ square from which we have removed a $2 \times 2$ square situated in one of its corners. We start by noticing that in any zone we need to have at least three red squares. Indeed, examining the two $3 \times 3$ squares that contain the two remaining opposite corners of the $5 \times 5$ square, they contain each at least two red squares and they only share one unit square (the one in the center), hence they contain together at least 3 red squares. With this remark, we move on to the promised induction:

For $a = 1$: we can paint red the squares situated in column number 3, on the lines 1, 3 and 4. Any $3 \times 3$ square contains exactly two of these red squares.

Assuming the assertion true for a $(3a - 1) \times (3a - 1)$ board, let us prove it for the $(3a+2) \times (3a+2)$ board. From the inductive hypothesis, in the $(3a-1) \times (3a-1)$ square situated in top-left corner we must have at least $2(a-1)^2 + a - 1$ red squares. We cover the remaining part of the $(3a+2) \times (3a+2)$ square with $3 \times 3$ squares, plus one zone placed in the bottom-right corner. In this remaining part we have at least $2(a-1)+2(a-1)+3$ red squares, hence we have in total at least $2(a-1)^2 + a - 1 + 4(a-1) + 3 = 2a^2 + a$ red squares.

For $m < n$ (the case when $m > n$ is similar), it is sufficient to notice that the rectangle $(3a + 2) \times (3b + 2)$ can be obtained from the $(3a + 2) \times (3a + 2)$ square by gluing a $(3a + 2) \times (3b - 3a)$ rectangle next to it. In the square there are at least $2a^2 + a$ red squares, while in the rectangle one can fit $a \cdot (b-a)$ disjoint $3 \times 3$ squares, therefore it contains at least $2a(b-a)$ red squares. In total, the rectangle has at least $2a^2 + a + 2a(b-a) = 2ab + a$ red squares. We have already seen that this number can actually be achieved, therefore the statement is proven.