The multiple of it's distance to the rest?

Vlad-3-14 wrote: The multiple of it's distance to the rest? I guess my terminology was not very accurate. Lets name the points $A_1,A_2,\dots, A_{100}$. I am trying to describe $|A_iA_1|.|A_iA_2|.\dots.|A_iA_{i-1}|.|A_iA_{i+1}|.\dots |A_iA_{100}|$ where $|AB|$ indicates the length of the segment $AB$.

The problem heavily relies on the following theorem: (McDougall) Let $n$ be a positive integer and $P_{i}$, for $1\leq i\leq 2n$ be cyclically ordered points on a circle. If $R_{i} = \prod_{\substack{1\leq j\leq 2n \\ j\neq i}} P_{i}P_{j}$, then \[\sum\limits_{i=1}^{n} \frac{1}{R_{2i}} = \sum\limits_{i=1}^{n} \frac{1}{R_{2i-1}}.\]Hence if we assume that such a set of $100$ points exists, then we can split the numbers $\{1,2,\dots, 100\}$ into two sets $A$ and $B$ of size $50$ such that $\sum\limits_{a\in A} \frac{1}{a} = \sum\limits_{b \in B} \frac{1}{b}$. However, this is clearly impossible, for example by considering $v_{97}$ of both sides, so the answer is no.