Note that a line segment of length $1$ has expected projection length equal to $\frac{2}{\pi}$ on a random line $l$. Thus by an averaging argument, there exists some line $l$ such that the projection lengths of $A_{2i-1}A_{2i}$ add up to at least the RHS quantity. Now suppose that the points $A_1 \sim A_{2n}$ project to $X_1 \sim X_{2n}$ on $l$, where $X_1 \sim X_{2n}$ are listed from left to right, and $X_i$ need not correspond to $A_i$. Then the sum of the projection lengths of $A_{2i-1}A_{2i}$ is at most $\sum_{i = 1}^{n} |X_i X_{n+i}| = \sum_{i = 1}^{n} |X_i X_{n+\tau(i)}|$ for any permutation $\tau$ of $\{1,2,\dots,n\}$. Let $C_1, C_2, \dots, C_n$ be the points in $A_1 \sim A_n$ that correspond to the projections $X_1, \dots, X_n$ respectively. Let $D_1, D_2, \dots, D_n$ be the remaining points (that correspond to projections $X_{n+1}, \dots, X_{2n}$). Then for any permutation $\tau$ we have $$ \sum_{i=1}^{n} |C_iD_{\tau(i)}| \geq \sum_{i=1}^{n} |X_i X_{n+ \tau(i)}| \geq \frac{2}{\pi} \cdot \sum_{i=1}^{n} |A_{2i-1}A_{2i}|.$$ It thus suffices to find a permutation $\tau$ such that the line segments $C_iD_{\tau(i)}$ are non-intersecting. We claim that such a matching between the "$C$-points" and the "$D$-points" exists whenever there is a line $q$ separating the two groups of points. Here we can take $q$ to be a line perpendicular to $l$ separating $X_{n}$ from $X_{n+1}$. To prove the claim, pick the line segment $C_iD_j$ whose intersection with $q$ is leftmost. Then all the other points $C_{i'}, D_{j'}$ must lie toward the right of the line determined by $C_iD_j$. This means $C_iD_j$ will not intersect with any other $C_{i'} D_{j'}$, so we can safely "match" these two points. Repeating the argument for the remaining points establishes the previous claim, and the problem is solved!

angiland wrote: We claim that such a matching between the "$C$-points" and the "$D$-points" exists whenever there is a line $q$ separating the two groups of points. Actually, your claim holds without a separating line. Just take the matching where the sum of the lengths become minimal, then one can show by triangle inequality that no segments intersect.

Oh yes, that's perfect

I know I’m late, but this was a beautiful solution to this hard problem! Can you share some of the motivation?