Consider any equilateral triangle $A_1A_2A_3$ inscribed in the given circle. It is well known (van Schooten's theorem; one-line proof using Ptolemy's identity) that for any point $P$ on the circle there exists a permutation $(i,j,k)$ of $(1,2,3)$ such that $|PA_i| = |PA_j| + |PA_k|$. This means $\sum_{\ell = 1}^3 |PA_{\ell}| = 2|PA_i| \leq 4$. Then, for $P_n$ being our points, $1\leq n \leq 60$, we have $60\cdot 4 \geq \sum_{n=1}^{60} \sum_{\ell = 1}^3 |P_nA_{\ell}| = \sum_{\ell=1}^{3} \sum_{n = 1}^{60} |P_nA_{\ell}|$, hence (at least) for one index $\ell \in \{1,2,3\}$ we have $\sum_{n = 1}^{60} |P_nA_{\ell}| \leq \dfrac {1} {3} \cdot 60\cdot 4 = 80$.