Tilings III

Colour the board like this:

Each $1\times 4$ tile covers exactly one blue square. We would need $10^2 \div 4=25$ tiles; but there are $26$ blue squares.

Label the spaces in the first row A, B, A, B, etc. Label the second row C, D, C, D, etc. Alternate the rows, so that each column only contains two alternating letters as well (A and C or B and D). Each 4x1 piece you place onto the grid covers pairs of letters - 2 A’s and 2 B’s, 2 B’s and 2 D’s, etc. Since the grid has 100 spaces, divided evenly into four letters, there are 25 of each, so they cannot all be covered working 2 at a time. Best case scenario, after 24 pieces are placed, you will be left with 1 A, 1 B, 1 C, and 1 D, which clearly cant all be covered by one piece.