We have a $6 \times 6$ square, partitioned into 36 unit squares. We select some of these unit squares and draw some of their diagonals, subject to the condition that no two diagonals we draw have any common points. What is the maximal number of diagonals that we can draw?
To solve the problem of finding the maximal number of diagonals that can be drawn in a $6 \times 6$ grid of unit squares such that no two diagonals intersect, we can use a combinatorial approach and a coloring argument.

1. **Understanding the Grid and Diagonals**:
   - A $6 \times 6$ grid consists of 36 unit squares.
   - Each unit square has two possible diagonals: one from the top-left to the bottom-right (let's call it a forward diagonal) and one from the top-right to the bottom-left (let's call it a backward diagonal).

2. **Non-Intersecting Condition**:
   - No two diagonals can intersect. This means that if a diagonal is drawn in one unit square, no diagonal can be drawn in any adjacent unit square that shares a vertex with the first diagonal.

3. **Coloring Argument**:
   - Consider coloring the $6 \times 6$ grid in a checkerboard pattern with alternating black and white squares.
   - In this pattern, each diagonal will connect squares of different colors.
   - Since no two diagonals can intersect, we can only draw diagonals in either all black squares or all white squares, but not both.

4. **Counting the Squares**:
   - In a $6 \times 6$ grid, there are 18 black squares and 18 white squares.
   - We can draw one diagonal in each of these squares without any intersections.

5. **Maximal Number of Diagonals**:
   - Since we can draw one diagonal in each of the 18 black squares (or alternatively, in each of the 18 white squares), the maximal number of non-intersecting diagonals is 18.

Therefore, the maximal number of diagonals that can be drawn in a $6 \times 6$ grid of unit squares, such that no two diagonals intersect, is 18.

The final answer is $\boxed{18}$