**UPDATE.** As pointed out in the comments, an upper bound exists for $x_i$ only if $\sum_{j=1}^{i-1} \frac{1}{x_j} < B$. If we have $B\leq \sum_{j=1}^{i-1} \frac{1}{x_j}<A$, then any sufficiently large $x_i,x_{i+1},\dots,x_n$ will work fine.

First, notice that the theorem is rather trivial. Indeed, for any $i=1,2,\dots,n$, we have:
$$R = \sum_{j=1}^n \frac{1}{x_j} \leq \sum_{j=1}^{i-1} \frac{1}{x_j} + \sum_{j=i}^{n} \frac{1}{x_i} = \sum_{j=1}^{i-1} \frac{1}{x_j} + (n-i+1)\frac{1}{x_i},$$
implying that
$$x_i \leq \frac{n-i+1}{R-\sum_{j=1}^{i-1} \frac{1}{x_j}}.$$

Replacing "$R = \sum_{j=1}^n \frac{1}{x_j}$" with "$B \leq \sum_{j=1}^n \frac{1}{x_j}$", we get a similar upper bound for $x_i$ (assuming that $\sum_{j=1}^{i-1} \frac{1}{x_j} < B$):
$$x_i \leq \frac{n-i+1}{B-\sum_{j=1}^{i-1} \frac{1}{x_j}}.$$

The bound is tight: it is attained when $\sum_{j=1}^n \frac{1}{x_j} = B$ and $x_i=x_{i+1}=\dots=x_n$.

The value of $A$ can be used to obtain a lower bound for $x_i$. Namely, from
$$A\geq \sum_{j=1}^n \frac{1}{x_j} \geq \sum_{j=1}^i \frac{1}{x_j}$$
we get
$$x_i \geq \frac{1}{A - \sum_{j=1}^{i-1} \frac{1}{x_j}}.$$

P.S. There is a related Kellogg problem, which for the case $R=1$ bounds $x_n$ *uniformly* as $x_n \leq s_{n-1}-1$ (where $s_j$ is Sylvester's sequence). This bound can be also extended to lower terms, giving $x_i \leq (n+1-i)\cdot (s_{i-1}-1)$ for all $i=1,2,\dots,n$.