【Analysis】This problem differs from regular fill-in questions in that it requires placing several "tetrominoes" so that there is no space left to place any more "tetrominoes".

【Solution】Solution: This problem requires the space to be utilized as "unreasonably" as possible, using the fewest "tetrominoes" to occupy the space, 
so that the remaining blank squares cannot accommodate any more "tetrominoes".
As shown in the figure below, after placing 3, there is no more space to place any "tetrominoes". If only 2 are placed, there will still be $25-2 \times 4=17$ squares left, which would inevitably leave a continuous space to place a "tetromino".

Therefore: The minimum number of "tetrominoes" to place so that no more can be placed is 3.
14. (15 points) There are five types of "tetromino" paper pieces as shown in the left figure, with an ample supply of each type. You need to place "tetrominoes" on a $5 \times 5$ grid as shown in the right figure. "Tetrominoes" can be flipped, and each small square of a "tetromino" must overlap with a small square of the grid. No two "tetrominoes" can overlap. What is the minimum number of "tetrominoes" that can be placed so that no more can be added?