A square is divided into $25$ unit squares by drawing lines parallel to the sides of the square. Some diagonals of unit squares are drawn from such that two diagonals do not share points. What is the maximum number diagonals that can be drawn with this property?
1. **Understanding the Problem:**
   - We have a large square divided into \(25\) unit squares, forming a \(5 \times 5\) grid.
   - We need to draw diagonals in such a way that no two diagonals share any points.

2. **Analyzing the Grid:**
   - Each unit square can have two possible diagonals: one from the top-left to bottom-right (denoted as \(\backslash\)) and one from the top-right to bottom-left (denoted as \(/\)).
   - Since the grid is \(5 \times 5\), there are \(25\) unit squares in total.

3. **Constraints:**
   - No two diagonals can share points. This means that if a diagonal is drawn in one unit square, no other diagonal can be drawn in any unit square that shares a vertex with it.

4. **Strategy:**
   - To maximize the number of diagonals, we need to ensure that the diagonals are placed in such a way that they do not interfere with each other.
   - One effective way to do this is to place diagonals in a checkerboard pattern. This ensures that no two diagonals share a vertex.

5. **Checkerboard Pattern:**
   - In a \(5 \times 5\) grid, a checkerboard pattern will have diagonals in alternating squares.
   - This pattern will cover half of the squares, as each diagonal occupies one square and leaves the adjacent squares free.

6. **Calculation:**
   - In a \(5 \times 5\) grid, there are \(25\) squares.
   - A checkerboard pattern will cover approximately half of these squares.
   - Therefore, the maximum number of diagonals that can be drawn without sharing points is \(\left\lfloor \frac{25}{2} \right\rfloor = 12.5\).

7. **Adjusting for Integer Count:**
   - Since we cannot have half a diagonal, we round down to the nearest whole number.
   - Thus, the maximum number of diagonals is \(12\).

The final answer is \(\boxed{12}\).