I think I have some sort of understanding of how to solve this but I'm not sure.
I would colour the board with 4 colours such that every 1×4 rectangle would cover one of each colour.
Then cover the bottom 4 rows such that we are left with a 2×6 rectangle.
Then fill the 4 leftmost columns so we are left with a 2×2 square which will have 2 tiles of the same colour hence it is impossible to cover the board with 1×4 tiles.
Would that be a correct answer?
Edit:
My colouring scheme works like this:
The 4 colours will be represented as 0,1,2 and 3.
Working along the top row of the board, we may as well label the first four squares 0, 1, 2 and 3, in that order.
In order to satisfy the property that any 4 × 1 rectangle placed on the board occupies one square of each colour, the next square along must be labelled 0. And the next square along must be labelled 1, and the next square along must be labelled 2, and the next square along must be labelled 3, and so on.
So we see that every square in the first row will be coloured according to the
repeating pattern 0, 1, 2, 3, 0, 1, 2, 3, . . .. Since this seems to work along the first row, we can use the same trick to fill the first column.
Then cycle through the rows marking tiles in the same pattern.
(The first tile of the second row is 1 so the tile to the left of it is 2, the next one is 3 and so on for every row)
Finally the board will be coloured in such a way that the square in
row i and column j is coloured i + j modulo 4.
Best Answer
Color the board like this: $$\matrix{1&2&3&4&1&2\cr2&3&4&1&2&3\cr3&4&1&2&3&4\cr4&1&2&3&4&1\cr1&2&3&4&1&2\cr2&3&4&1&2&3\cr}$$ Show that wherever you put a $4\times1$ it covers one of each color, note that there are 10 2s and only 8 4s to be covered, QED.