A convex 2011-gon is drawn on the board. Peter keeps drawing its diagonals in such a way, that each newly drawn diagonal intersected no more than one of the already drawn diagonals. What is the greatest number of diagonals that Peter can draw?
To solve the problem, we need to determine the greatest number of diagonals that Peter can draw in a convex 2011-gon such that each newly drawn diagonal intersects no more than one of the already drawn diagonals. We claim that the answer is \(2n - 6\) for any \(n\)-gon. We will prove this bound and its constructability through strong induction.

1. **Base Case:**
   - For \(n = 3\) (a triangle), there are no diagonals, so the number of diagonals is \(0\), which satisfies \(2 \cdot 3 - 6 = 0\).
   - For \(n = 4\) (a quadrilateral), there are \(2\) diagonals, which satisfies \(2 \cdot 4 - 6 = 2\).

2. **Inductive Hypothesis:**
   - Assume that for any \(k\)-gon where \(3 \leq k \leq n-1\), the maximum number of diagonals that can be drawn such that each newly drawn diagonal intersects no more than one of the already drawn diagonals is \(2k - 6\).

3. **Inductive Step:**
   - Consider an \(n\)-gon. We need to show that the maximum number of diagonals that can be drawn is \(2n - 6\).
   - The last diagonal drawn divides the \(n\)-gon into an \(m\)-gon and a \((n+2-m)\)-gon. This final diagonal must intersect with at most one other diagonal.
   - By the inductive hypothesis, the maximum number of diagonals in the \(m\)-gon is \(2m - 6\), and the maximum number of diagonals in the \((n+2-m)\)-gon is \(2(n+2-m) - 6\).
   - Therefore, the total number of diagonals is:
     \[
     (2m - 6) + (2(n+2-m) - 6) + 1 = 2m - 6 + 2n - 4 - 2m + 1 - 6 = 2n - 6
     \]
     The "+1" term accounts for the final diagonal that intersects with at most one other diagonal.

4. **Constructability:**
   - To achieve this bound, consider four consecutive vertices \(A, B, C, D\) in the \(n\)-gon. Draw the diagonal \(AC\) as the final diagonal and \(BD\) as the only diagonal crossing \(AC\).
   - This leaves us with two smaller polygons: one with \(m\) vertices and the other with \(n+2-m\) vertices.
   - By the inductive hypothesis, we can draw \(2(m-1) - 6\) diagonals in the \(m\)-gon and \(2((n+2-m)-1) - 6\) diagonals in the \((n+2-m)\)-gon.
   - Adding these diagonals and the final diagonal \(AC\), we get:
     \[
     2(m-1) - 6 + 2((n+2-m)-1) - 6 + 1 = 2n - 6
     \]
   - Thus, the maximum number of diagonals that can be drawn in an \(n\)-gon is \(2n - 6\).

The final answer is \(\boxed{2n - 6}\).