16. (22 points) Arrange $m$ guests of the same gender to stay in $A_{1}, A_{2}, \cdots, A_{n}$, these $n$ rooms as follows: First, arrange 1 guest and $\frac{1}{7}$ of the remaining guests in room $A_{1}$; then, from the remaining guests, arrange 2 guests and $\frac{1}{7}$ of the remaining guests in room $A_{2}$; and so on, the number of guests in the $i$-th room is $i$ plus $\frac{1}{7}$ of the remaining guests. In this way, the last room $A_{n}$ is exactly arranged with the last remaining $n$ guests. Try to find the number of guests and the number of rooms, as well as the number of guests in each room.
16. Let the number of guests remaining after arranging the $k$-th room $A_{k}$ be $a_{k}$, then $a_{0}=m, a_{n-1}=n$.

Since the number of guests staying in the $k$-th room $A_{k}$ is $k + \frac{a_{k-1}-k}{7}$, we have $a_{k-1}-a_{k}=k+\frac{a_{k-1}-k}{7}$, which simplifies to
$a_{k}=\frac{6}{7}\left(a_{k-1}-k\right)$.
Rearranging gives $a_{k}+6 k-36=\frac{6}{7}\left[a_{k-1}+6(k-1)-36\right]$.
This indicates that the sequence $b_{k}=a_{k}+6 k-36$ is a geometric sequence with a common ratio of $q=\frac{6}{7}$, where $b_{0}=a_{0}-36=m-36, b_{n-1}=a_{n-1}+6(n-1)-36=7 n-42$.

Substituting into the general term formula gives $7 n-42=(m-36)\left(\frac{6}{7}\right)^{n-1}$, which simplifies to $m=36+\frac{7^{n}(n-6)}{6^{n-1}}$.
Since $m$ is a positive integer and $7^{n}$ and $6^{n-1}$ are coprime, it follows that $6^{n-1} \mid (n-6)$.
However, $0 \leqslant\left|\frac{n-6}{6^{n-1}}\right|<1$ for $n>1$, solving this yields $n=6$.
Thus, $m=36$.
Therefore, the $A_{1}$ room accommodates $1+\frac{36-1}{7}=6$ guests; the $A_{2}$ room accommodates $2+\frac{30-2}{7}=6$ guests; the $A_{3}$ room accommodates $3+\frac{24-3}{7}=6$ guests; the $A_{4}$ room accommodates $4+\frac{18-4}{7}=6$ guests; the $A_{5}$ room accommodates $5+\frac{12-5}{7}=6$ guests; and the last room accommodates the remaining 6 guests.

In summary, there are 36 guests in total, 6 rooms, and each room accommodates 6 guests.