Answer: 1011.
We show that the answer for $2n$ points is $n$. Let us numerate marked points in clockwise order by $1, 2, \dots, 2n$. The arc starting at point $a$ and ending at $b$ in clockwise order will be denoted by $[a, b]$.

The $n$ arcs chosen as $[1, 2], [2, 4], [3, 6], [4, 8], \dots, [n-1, 2n-2], [n, 2n]$ satisfy conditions.

Now suppose that $k \ge n+1$ chosen arcs satisfy problem conditions. Let the shortest and longest chosen arcs are $[a, b]$ and $[c, d]$, respectively. Suppose that the arc lengths of $[a, b]$ and $[c, d]$ are $l$ and $m$, respectively.

Case 1. $[a, b] \cap [c, d] = \emptyset$.
Any chosen arc except $[a, b]$ and $[c, d]$ will either start on arc $[a+1, c-1]$ or end on arc $[d+1, b-1]$. Internal marked points of $[a, b]$ can be both starting and ending points of chosen arcs. Therefore, the sum of the number of possible starting and ending points is $t = 2n - (m+1) + (l-1)$. Since $m \ge l+n$ we get $t \le n-2$. There are at least $n-1$ chosen arcs except $[a, b]$ and $[c, d]$. Each of them either starts or ends at one of these $n-2$ points. Therefore, two of these arcs either starts or ends at the same marked point, a contradiction.

Case 2. $[a, b] \cap [c, d] \neq \emptyset$. Without loss of generality $a \in (c, d]$.
Any chosen arc except $[a, b]$ and $[c, d]$ will either will start on arc $[a+1, c-1]$ or end on arc $[d+1, b-1]$. Let $k$ be the length of arc $[d, b]$. Internal marked points of $[d, b]$ can be both starting and ending points of chosen arcs. Therefore, the sum of the number of possible starting and ending points is $t = 2n - (m+1) + (l-1)$. Since $m \ge l+n$ we get $t \le n-2$. There are at least $n-1$ chosen arcs except $[a, b]$ and $[c, d]$. Each of them either starts or ends at one of these $n-2$ points. Therefore, two of these arcs either starts or ends at the same marked point, a contradiction. Done.