What is the sum of all $k\leq25$ such that one can completely cover a $k\times k$ square with $T$ tetrominos (shown in the diagram below) without any overlap?

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$\text{(A) }20\qquad\text{(B) }24\qquad\text{(C) }84\qquad\text{(D) }108\qquad\text{(E) }154$
1. **Understanding the problem**: We need to find the sum of all \( k \leq 25 \) such that a \( k \times k \) square can be completely covered with \( T \) tetrominos without any overlap. A tetromino is a shape made up of 4 squares.

2. **Analyzing the tetromino**: Each tetromino covers exactly 4 unit squares. Therefore, for a \( k \times k \) square to be completely covered by tetrominos, the total number of unit squares in the square, \( k^2 \), must be divisible by 4. This implies:
   \[
   k^2 \equiv 0 \pmod{4}
   \]
   Since \( k^2 \equiv 0 \pmod{4} \), \( k \) itself must be divisible by 2. However, we need to check further conditions to ensure complete coverage.

3. **Checking for \( k \) values**: We need to check which values of \( k \) (where \( k \leq 25 \)) satisfy the condition \( k^2 \equiv 0 \pmod{4} \). These values are:
   \[
   k = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
   \]
   However, we need to ensure that these values can indeed be covered by tetrominos without any gaps or overlaps.

4. **Verifying the coverage**: For a \( k \times k \) square to be completely covered by tetrominos, \( k \) must be a multiple of 4. This is because a \( 2 \times 2 \) square cannot be covered by tetrominos without leaving gaps. Therefore, we need to consider only those \( k \) which are multiples of 4:
   \[
   k = 4, 8, 12, 16, 20, 24
   \]

5. **Summing the valid \( k \) values**: We sum the values of \( k \) that are multiples of 4 and less than or equal to 25:
   \[
   4 + 8 + 12 + 16 + 20 + 24 = 84
   \]

Conclusion:
\[
\boxed{84}
\]