A [i]triminó[/i] is a rectangular tile of $1\times 3$. Is it possible to cover a $8\times8$ chessboard using $21$ triminós, in such a way there remains exactly one $1\times 1$ square without covering? In case the answer is in the affirmative, determine all the possible locations of such a unit square in the chessboard.
1. **Understanding the Problem:**
   We need to determine if it is possible to cover an $8 \times 8$ chessboard using $21$ triminós ($1 \times 3$ tiles) such that exactly one $1 \times 1$ square remains uncovered. Each triminó covers exactly $3$ squares.

2. **Total Number of Squares:**
   The total number of squares on an $8 \times 8$ chessboard is:
   \[
   8 \times 8 = 64
   \]
   If we use $21$ triminós, the number of squares they cover is:
   \[
   21 \times 3 = 63
   \]
   This leaves exactly one square uncovered, which matches the problem's requirement.

3. **Divisibility by 3:**
   To determine the possible locations of the uncovered square, we need to consider the properties of the triminós. Each triminó covers three consecutive squares, and the sum of the indices of these squares is divisible by $3$.

4. **Labeling the Chessboard:**
   We can label the squares of the chessboard with numbers $1, 2, 3$ in a repeating pattern:
   \[
   \begin{array}{|c|c|c|c|c|c|c|c|}
   \hline
   1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 \\
   \hline
   1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 \\
   \hline
   1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 \\
   \hline
   1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 \\
   \hline
   1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 \\
   \hline
   1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 \\
   \hline
   1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 \\
   \hline
   1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 \\
   \hline
   \end{array}
   \]

5. **Sum of All Numbers:**
   The sum of all numbers on the chessboard is:
   \[
   21 \times 1 + 21 \times 2 + 22 \times 3 = 21 + 42 + 63 = 126
   \]
   Since $126$ is divisible by $3$, the sum of the numbers covered by the triminós must also be divisible by $3$.

6. **Uncovered Square:**
   The uncovered square must be in a position such that the sum of the numbers covered by the triminós remains divisible by $3$. This means the uncovered square must be in a position labeled with a number that, when removed, still leaves the sum divisible by $3$.

7. **Possible Locations:**
   Since the sum of all numbers is $126$, which is divisible by $3$, the uncovered square must be in a position labeled with a number that, when removed, still leaves the sum divisible by $3$. Therefore, the uncovered square must be in a position labeled with $3$.

8. **Conclusion:**
   The uncovered square can be in any position labeled with $3$ on the chessboard. These positions are:
   \[
   (1, 3), (1, 6), (2, 3), (2, 6), (3, 3), (3, 6), (4, 3), (4, 6), (5, 3), (5, 6), (6, 3), (6, 6), (7, 3), (7, 6), (8, 3), (8, 6)
   \]

The final answer is \( \boxed{ (1, 3), (1, 6), (2, 3), (2, 6), (3, 3), (3, 6), (4, 3), (4, 6), (5, 3), (5, 6), (6, 3), (6, 6), (7, 3), (7, 6), (8, 3), (8, 6) } \)