Solution:
We provide the following construction, which shows that the answer is at most four. Clearly, four T-tetrominoes tile a $4 \times 4$ grid, as follows:

These can be used to tile the upper-left $200 \times 200$ grid. It can be verified that the following pattern fills the remaining space, leaving four grid squares uncovered:

As the answer must be a multiple of four, it remains to prove that the answer is not zero, that is, it is impossible to tile a $202 \times 202$ grid with T-tetrominoes.

Number the rows of the grid, from left to right, with $0,1, \ldots, 201$. Similarly, number the columns of the grid, from top to bottom, with $0,1, \ldots, 201$.
Assign the square numbered $(x, y)$ with the weight $4x+4y+1$. It can be verified that a T-tetromino, no matter how it is placed, covers a sum of weights divisible by $8$. However, the sum of all the weights is not divisible by $8$, as desired.

Solution:
An alternative way to prove that the $202 \times 202$ grid cannot be tiled with T-tetrominoes is with a checkerboard coloring.

For the sake of contradiction, assume that the board can be tiled with T-tetrominoes. Clearly, a T-tetromino covers only either three or one black square. Let $x$ be the number of T-tetrominoes each covering three black squares and $y$ be the number of T-tetrominoes each covering only one black square. Counting the number of black and white squares yields
$$3x + y = 2 \cdot 101^{2}, \quad x + 3y = 2 \cdot 101^{2} \quad \Longrightarrow \quad x = y = \frac{101^{2}}{2}$$
which contradicts the fact that $x$ and $y$ are integers.