We will show that it is possible iff $n$ is even (and $n\geq 3$). First, we will prove that it is impossible for odd $n$. Construct a graph where each chord is a vertex and two vertices are connected iff the chords intersect. We know that each vertex has degree $n-2$, and there are $n$ vertices. Therefore, the sum of degrees of all vertices is $n(n-2)$. However, for odd $n$, this is an odd number, but it is also twice as big as the number of edges and therefore even, which is a contradiction. So no odd $n$ work. To see that it is possible for all even $n$, just take $n/2$ pairs of chords parallel to each other and very close to the diameter parallel to them.

We will show that it is possible for all $n$ ($\geq 3$). If $n$ is divisible by $3$, just take $n/3$ triplets of parallel lines close to the diameter parallel to them. If $n$ is congruent to $1$ mod $3$, do the same as above, but replace one of the triplets with a quadruplet of lines which are all really close to being parallel to each other and some diameter, form $2$ pairs where each line intersects the other in the pair, and are really close to the diameter. If $n$ is congruent to $2$ mod $3$ and greater than $5$, do the same as above, but with $2$ quadruplets instead of $1$. And for $n=5$, form a regular pentagon with all vertices inside the circle, but really close to the boundary, and extend the sides.