Let $AB$ and $CD$ be the diameters of the circle which are parallel and perpendicular to $L$ respectively. Let $P_iQ_i$ be the projection of each segment on $AB$, and let $X_iY_i$ be the projection of each segment on $CD$. Note that $P_iQ_i + X_iY_i \ge 1$. Therefore, we have
$$\sum_{j=1}^{4n} P_iQ_i + \sum_{j=1}^{4n} X_iY_i = \sum_{j=1}^{4n} (P_iQ_j + X_iY_j) \ge 4n = AB + CD.$$
WLOG assume $\sum_{j=1}^{4n} P_iQ_i \ge AB$. As all the segments $P_iQ_i$ lie strictly inside the segment $AB$, two of these segments $P_iQ_i$ and $P_jQ_j$ must overlap. Let $E$ be a point lying on both of them. Then the line passing through $E$ and perpendicular to $AB$ intersects two unit segments whose projections are $P_iQ_i$ and $P_jQ_j$. This completes the proof.