If $k \nmid 2022$, start tiling the grid with $k \times k$ sized tiles from the top left, so there are $\left \lfloor \frac{2022}{k} \right \rfloor ^2$ tiles. Whenever Bob colours a $k \times k$ square blue, exactly one bottom right corner square of a tile gets coloured, so atmost $\left(\frac{\left \lfloor \frac{2022}{k} \right \rfloor ^2}{2}\right)$ $k\times k$ tiles are coloured giving a bound of $k\left(\frac{\left \lfloor \frac{2022}{k} \right \rfloor ^2}{2}\right)$ blue squares which is $< 2022^2/2$ so Alice can colour the rest red and win. If $k$ is an even divisor of $2022$ then $2022/k$ is odd, start tiling the grid again as above, with a similar argument Alice wins (because Alice goes first). If $k$ is an odd divisor of $2022$ then $2022/k$ is even, start tiling similarly, this time note that the number of tiles is even so the best Bob can do is colour an entire white tile blue at every move, in the end half the tiles are blue and Alice colours the rest of the squares red to secure a draw.