Given a regular $n$-sided polygon with $n \ge 3$. Maria draws some of its diagonals in such a way that each diagonal intersects at most one of the other diagonals drawn in the interior of the polygon. Determine the maximum number of diagonals that Maria can draw in such a way.

Note: Two diagonals can share a vertex of the polygon. Vertices are not part of the interior of the polygon.
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   - For an \( n \)-sided polygon, we can pair the vertices in such a way that each pair forms a diagonal, and these diagonals do not intersect with more than one other diagonal.
   - For example, in a hexagon (\( n = 6 \)), we can draw the diagonals (1,4), (2,5), and (3,6). Each of these diagonals intersects exactly one other diagonal.

6. **Generalizing the Pattern:**
   - For an even \( n \), we can pair the vertices as described above.
   - For an odd \( n \), we can pair \( n-1 \) vertices and leave one vertex out, ensuring that the condition is still satisfied.

7. **Counting the Maximum Number of Diagonals:**
   - For an even \( n \), the maximum number of diagonals is:
     \[
     \frac{n}{2}
     \]
   - For an odd \( n \), the maximum number of diagonals is:
     \[
     \frac{n-1}{2}
     \]

8. **Verification:**
   - We need to verify that this construction satisfies the condition that each diagonal intersects at most one other diagonal.
   - By pairing the vertices as described, each diagonal intersects exactly one other diagonal, satisfying the condition.

Conclusion:
The maximum number of diagonals that Maria can draw in an \( n \)-sided polygon such that each diagonal intersects at most one other diagonal is:
\[
\boxed{\left\lfloor \frac{n}{2} \right\rfloor}
\]