Solution I. We can place the Residents No. $1,2, \ldots$ according to the following rule. First, put Resident No. 1 arbitrarily. For $n>2$, if Residents No. $1,2, \ldots, n-1$ have already been placed, we consider the positions where Resident No. n cannot be placed. For $1 \leq m \leq n-1$, by $d_{m, n} \geq \frac{1}{m+n}$, we know that the Resident No. $n$ cannot be placed in the arc that is centered at Resident No. $m$, and of the length $\frac{2}{m+n}$. The total length of these arcs is $\frac{2}{n+1}+\frac{2}{n+2}+\cdots+\frac{2}{2 n-1}<2\left(\ln \frac{n+1}{n}+\ln \frac{n+2}{n+1}+\cdots+\ln \frac{2 n-1}{2 n-2}\right)=2 \ln \frac{2 n-1}{n}<2 \ln 2$. Therefore, the total length of the union of these arcs does not exceed $2 \ln 2$, while the perimeter of the circle is $\frac{1}{4} \cdot 2 \pi=\frac{\pi}{2}$. It is easy to observe that $\frac{\pi}{2}>1.5>2 \ln 2$, so these arcs would not cover the whole circle, hence it is always possible to find a place for Resident No. $n$ such that its distances to Residents No. $1,2, \ldots, n-1$ satisfy the requirement. By induction we conclude that the circle can accommodate any quantity of residents.