Given an integer number $n \geq 3$, consider $n$ distinct points on a circle, labelled $1$ through $n$. 
Determine the maximum number of closed chords $[ij]$, $i \neq j$, having pairwise non-empty intersections. 

[i]János Pach[/i]
1. **Restate the problem**: We are given \( n \geq 3 \) distinct points on a circle, labeled \( 1 \) through \( n \). We need to determine the maximum number of closed chords \([ij]\), \( i \neq j \), such that all chords have pairwise non-empty intersections.

2. **Fix an orientation and relabel the points**: Fix an orientation of the circle and relabel the points \( 1 \) through \( n \) in the corresponding circular order.

3. **Consider a configuration of chords**: Consider a configuration of chords \([ij]\), \( i \neq j \), with pairwise non-empty intersections.

4. **Assign points to chords**: Assign to each point \( i \), which is an endpoint of at least one chord, the first point \( i' \) following \( i \) to which it is connected.

5. **Show that deleting certain chords leaves no chords**: We now show that by deleting the chords \([ii']\), no chord is left, so the number of chords in the configuration does not exceed \( n \).

6. **Proof by contradiction**: Suppose, if possible, that some chord \([ij]\) is left. Then \( i \), \( i' \), \( j \), \( j' \) are in circular order around the circle, so the chords \([ii']\) and \([jj']\) do not meet - a contradiction.

7. **Construct a maximal configuration**: A maximal configuration is given by the \( n \) chords \([1i]\), \( i = 2, 3, \ldots, n \), and \([2n]\).

8. **Conclusion**: Therefore, the maximum number of closed chords with pairwise non-empty intersections is \( n \).

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