Solution:

First, observe that
$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2} \quad \text{and} \quad \sum_{i=n+1}^{2n} i = n^{2} + \frac{n(n+1)}{2}$$
which means that
$$\sum_{i=n+1}^{2n} i - \sum_{i=1}^{n} i = n^{2}$$
Let $A = \{1, 2, \ldots, n\}$ and $B = \{n+1, n+2, \ldots, 2n\}$. (Here, we do not distinguish the point labeled $x$ and the number $x$ itself.) Because the numbers are arranged on a circle, one can find a pair $\{x_{1}, y_{1}\}$, where $x_{1} \in A$ and $y_{1} \in B$, such that one arc joining $x_{1}$ and $y_{1}$ contains no other labeled points. One can then remove the chord (including $x_{1}$ and $y_{1}$) joining these points. Among the remaining labeled points, one can find again a pair $\{x_{2}, y_{2}\}$, where $x_{2} \in A \setminus \{x_{1}\}$ and $y_{2} \in B \setminus \{y_{1}\}$, such that one arc joining $x_{2}$ and $y_{2}$ does not contain a labeled point, and then remove again the chord (including the endpoints) joining $x_{2}$ and $y_{2}$. Continuing this process, one can find pairs $\{x_{3}, y_{3}\}, \{x_{4}, y_{4}\}$, and so on, and then remove the chords joining the pairs.

We claim that the removed chords satisfy the required properties. Clearly, there are $n$ such chords. Because no labeled point lies on one arc joining $x_{j}$ and $y_{j}$ for any $1 \leq j \leq n$, the removed chords are non-intersecting. Finally, the sum of the values assigned to the removed chords is
$$\sum_{j=1}^{n} (y_{j} - x_{j}) = \sum_{j=1}^{n} y_{j} - \sum_{j=1}^{n} x_{j} = \sum_{i=n+1}^{2n} i - \sum_{i=1}^{n} i = n^{2}$$
This ends the proof of our claim.