$ 2n\plus{}1$ lines are drawn in the plane, in such a way that every 3 lines define a triangle with no right angles. What is the maximal possible number of acute triangles that can be made in this way?
1. **Understanding the Problem:**
   We are given \(2n + 1\) lines in the plane such that every three lines form a triangle with no right angles. We need to find the maximum number of acute triangles that can be formed.

2. **Simplifying the Problem:**
   The acuteness of a triangle formed by three lines depends on the angles between the lines. We can assume, without loss of generality, that all lines are tangents to a common circle. This is because the relative angles between the lines remain unchanged if we replace each line with a parallel line.

3. **Optimal Configuration:**
   The optimal configuration for maximizing the number of acute triangles is when the lines are the sides of a regular \((2n + 1)\)-gon. This is because in a regular polygon, the angles between adjacent sides are equal, and this symmetry helps in forming more acute triangles.

4. **Counting Acute Triangles:**
   In a regular \((2n + 1)\)-gon, the number of acute triangles can be calculated using combinatorial methods. Specifically, we need to count the number of sets of three lines that form acute triangles.

5. **Combinatorial Calculation:**
   The number of ways to choose 3 lines out of \(2n + 1\) lines is given by the binomial coefficient:
   \[
   \binom{2n + 1}{3}
   \]
   However, not all of these triangles will be acute. We need to determine the proportion of these triangles that are acute.

6. **Proportion of Acute Triangles:**
   For a regular \((2n + 1)\)-gon, the proportion of acute triangles can be derived from geometric properties. It turns out that the number of acute triangles is:
   \[
   \frac{1}{3}(2n + 1) \binom{n + 1}{2} = \frac{1}{4} \binom{2n + 2}{3}
   \]
   This formula accounts for the specific arrangement of the lines and the angles between them.

7. **Proportion Analysis:**
   The proportion of acute triangles among all possible triangles is:
   \[
   \frac{\frac{1}{4} \binom{2n + 2}{3}}{\binom{2n + 1}{3}} = \frac{n + 1}{2(2n - 1)}
   \]
   As \(n\) grows, this proportion approaches \(\frac{1}{4}\), which is approximately 25%.

8. **Conclusion:**
   The maximal possible number of acute triangles that can be formed by \(2n + 1\) lines in the plane, where every three lines form a triangle with no right angles, is given by:
   \[
   \frac{1}{4} \binom{2n + 2}{3}
   \]

The final answer is \( \boxed{ \frac{1}{4} \binom{2n + 2}{3} } \).