Suppose $n$ lines in plane are such that no two are parallel and no three are concurrent. For each two lines their angle is a real number in $[0,\frac{\pi}2]$. Find the largest value of the sum of the $\binom n2$ angles between line.
[i]By Aliakbar Daemi[/i]
1. **Initial Setup and Assumptions:**
   - We are given \( n \) lines in a plane such that no two lines are parallel and no three lines are concurrent.
   - For each pair of lines, the angle between them is a real number in \([0, \frac{\pi}{2}]\).
   - We aim to find the largest value of the sum of the \(\binom{n}{2}\) angles between the lines.

2. **Reduction to Concurrent Lines:**
   - Translating a line does not affect its angle with another line. Therefore, we can assume without loss of generality (WLOG) that all lines are concurrent at some point \( O \).

3. **Key Claim:**
   - The maximum sum of angles is attained when two of the lines are perpendicular.

4. **Proof of the Key Claim:**
   - Consider any line \(\ell\) in the set \( S \) of \( n \) lines.
   - Let \( m \) be the line passing through \( O \) and perpendicular to \(\ell\) (assuming \( m \) is not currently in \( S \)).
   - Assume all lines lie on the same side of \(\ell\) (say above \(\ell\)).
   - Let \(\ell_1, \ell_2, \ldots, \ell_{n-1}\) be the \( n-1 \) lines other than \(\ell\).
   - On the left and right sides of \( m \), there are lines. WLOG, assume the left side has \(\ge\) lines than the right side.
   - Let \(\ell_i\) be the line next to \( m \) on the left side. Replacing \(\ell_i\) by \( m \) does not decrease the sum of angles, proving our claim.

5. **Angle Sum Property:**
   - If two lines \( a \) and \( b \) are perpendicular, then for any line \( x \), we have:
     \[
     \angle(a, x) + \angle(b, x) = \frac{\pi}{2}
     \]

6. **Recurrence Relation:**
   - Let \( f(n) \) denote the maximum sum of angles for \( n \) lines.
   - Using our claim, we obtain the recurrence relation:
     \[
     f(n) = f(n-2) + (n-1) \cdot \frac{\pi}{2} \quad \forall \, n \ge 3
     \]

7. **Base Cases:**
   - \( f(1) = 0 \)
   - \( f(2) = \frac{\pi}{2} \)

8. **Solving the Recurrence:**
   - For odd \( n = 2k+1 \):
     \[
     f(2k+1) = f(1) + \frac{\pi}{2} \left(2k + (2k-2) + \cdots + 2 \right) = \frac{\pi}{2} \cdot k(k+1)
     \]
   - For even \( n = 2k+2 \):
     \[
     f(2k+2) = f(2) + \frac{\pi}{2} \left((2k+1) + (2k-1) + \cdots + 1 \right) = \frac{\pi}{2} \cdot (k+1)^2
     \]

9. **Conclusion:**
   - The maximum sum of the angles between the \( n \) lines is given by:
     \[
     f(n) = \begin{cases}
     \frac{\pi}{2} \cdot k(k+1) & \text{if } n = 2k+1 \\
     \frac{\pi}{2} \cdot (k+1)^2 & \text{if } n = 2k+2
     \end{cases}
     \]

The final answer is \(\boxed{\frac{\pi}{2} \cdot \left\lfloor \frac{n}{2} \right\rfloor \left(\left\lfloor \frac{n}{2} \right\rfloor + 1\right)}\).