Let $p$ be a prime number. Let $r$ be a number with $1 \leq r \leq \frac{p-3}{2}$. Let $T(p,r) \subset \mathbb Z^r$ be the set of all $r$-tuples $(n_1, n_2, \dotsc, n_r)\in \mathbb Z^r$ such that $1 < n_1 < n_2 < \dotsc < n_r < p-1$ and $n_i-n_{i-1} \geq 2$ for every $i \in \{2,\dotsc, r\}$.
Next, consider the positive rational number $S(p,r) \in \mathbb Q$ defined as following:
$$ S(p,r) = \sum_{T(p,r)}\frac{1}{n_1 \cdot n_2 \cdots n_r} = \sum_{\substack{1 < n_1 < n_2 < \dotsc < n_r < p-1 \\ n_i-n_{i-1} \geq 2}} \frac{1}{n_1 \cdot n_2 \cdots n_r}. $$
Finally, consider the field $\mathbb F_p$. Since $p$ does not divide $n_i$ whenever $1 < n_i < p-1$, each of the products $n_1 \cdot n_2 \cdots n_r$ appearing in any of the denominators of the sum $S(p,r)$ is non-zero modulo $p$, hence we can evaluate $S(p,r)$ in the field $\mathbb F_p$.
I have reasons to believe the following is true: for every prime number $p$ and every $r$ with $1 \leq r \leq \frac{p-3}{2}$,
\begin{equation} S(p,r) \equiv 0 \mod p. \end{equation}
The first reason that I think this equality in $\mathbb F_p$ should hold, is that the equality holds for $r = 2$ and for odd values of $r$, which is not too hard to show. For example, if $r$ is odd, then $(n_1, n_2, \dotsc, n_r) \in T(p,r)$ implies $(p-n_r, p-n_{r-1}, \dotsc, p-n_1) \in T(p,r)$; moreover, $(n_1\cdot n_2 \cdots n_r) + (p-n_r)\cdot(p-n_{r-1}) \cdots (p-n_1) \equiv 0 \mod p$.
The second reason that I think this is true, is that computer calculations show that $S(p,r) \equiv 0 \mod p$ for every $p$ and $r$ such that the prime number $p$ is smaller than $2000$.
Does anybody have an idea how to prove the equality in full generality? Or does anybody know if this (or something similar) has been done before, and can provide me with a reference?