Note that $|z_i-1|\leq r$ implies $|\arg z_i|\leq\cos^{-1}\sqrt{1-r^2}$. We shall generalise the inequality. Fix $0<\theta<90^\circ$; we shall show \[\left|\sum z_i\right|\left|\sum\frac1{z_i}\right|\geq n^2\cos^2\theta\] for nonzero complex numbers $z_1,\ldots,z_n$ with $|\arg z_i|\leq\theta$. This immediately implies the desired inequality. Now $\Re(z_i)=|z_i|\cos\arg z_i$ and $\Re\left(\frac1{z_i}\right)=\frac1{|z_i|}\cos\arg z_i$, so \[\Re(z_i)\Re\left(\frac1{z_i}\right)=\cos^2\arg z_i\geq\cos^2\theta.\] Therefore \begin{align*} \left|\sum z_i\right|\left|\sum\frac1{z_i}\right| &\geq\Re\left(\sum z_i\right)\Re\left(\sum\frac1{z_i}\right)\quad(\because |z|\geq\Re(z))\\ &=\left(\sum\Re(z_i)\right)\left(\sum\Re\left(\frac1{z_i}\right)\right)\\ &\geq\left(\sum\sqrt{\Re(z_i)\Re\left(\frac1{z_i}\right)}\right)^2\quad\mbox{(C-S)}\\ &\geq(n\cos\theta)^2, \end{align*} as desired.

Let $z_k=x_k+y_k{i} $, hence $(x_k-1)^2+y^2_k\le r^2$ . $ \left | \sum_{i=1}^n z_i \right | \cdot \left | \sum_{i=1}^n \frac{1}{z_i} \right |=\sqrt{( \sum_{k=1}^nx_k)^2+( \sum_{k=1}^ny_k)^2}\cdot\sqrt{( \sum_{k=1}^n\frac{x_k}{x^2_k+y^2_k})^2+( \sum_{k=1}^n\frac{y_k}{x^2_k+y^2_k})^2}$ $\ge \left | \sum_{k=1}^n x_k \right | \left | \sum_{k=1}^n\frac{x_k}{x^2_k+y^2_k} \right | +\left |\sum_{k=1}^ny_k \right | \left | \sum_{k=1}^n\frac{y_k}{x^2_k+y^2_k} \right |\ge \left | \sum_{k=1}^n\frac{x_k}{\sqrt{x^2_k+y^2_k}} \right | $, Hence, we need to prove that $ \frac{x_k}{\sqrt{x^2_k+y^2_k}}\ge \sqrt{1-r^2} \iff \left | \frac{y_k}{x_k} \right | \le \frac{r}{\sqrt{1-r^2}}.$

Setting $z_i=a_i+ib_i,$ where $a_i,\,b_i \in \mathbb R.$ From the given hypothesis, we have $$(1-a_i)^2+b_i^2 \le r^2,$$ or $$(a_i^2+b_i^2)+(1-r^2) \le 2a_i.$$ Using the AM-GM inequality, we have $2a_i \ge 2\sqrt{(1-r^2)(a_i^2+b_i^2)}.$ From this, it follows that $a_i>0$ and $$\frac{a_i^2}{a_i^2+b_i^2} \ge 1-r^2,\quad \forall i =1,\,2,\, \ldots,\, n.\quad (*)$$ Since $ \sum _{i=1}^n z_i=\sum _{i=1}^n a_i+i\sum_{i=1}^n b_i$ and $\sum_{i=1}^n \frac{1}{z_i}=\sum_{i=1}^{n} \frac{a_i-ib_i}{a_i^2+b_i^2},$ the original inequality can be written as $$\sqrt{\left(\sum_{i=1}^na_i\right)^2+\left(\sum_{i=1}^nb_i\right)^2}\cdot \sqrt{\left(\sum_{i=1}^n \frac{a_i}{a_i^2+b_i^2}\right)^2+\left(\sum_{i=1}^{n}\frac{b_i}{a_i^2+b_i^2}\right)^2} \ge n^2(1-r^2).$$ Using $(*),$ we get $$\mathrm{LHS} \ge \left(\sum_{i=1}^n a_i\right)\left(\sum_{i=1}^n \frac{a_i}{a_i^2+b_i^2}\right) \ge \left(\sum_{i=1}^n a_i\right)\left(\sum_{i=1}^n \frac{1-r^2}{a_i}\right) \ge n^2(1-r^2),$$ as desired.

Remarks: When $n$ is even, $n^2(1-r^2)$ is the minimum of $|z_1+z_2+\cdots+z_n|\cdot |1/z_1+1/z_2+\cdots+1/z_n|$. However, when $n$ is odd, no. Some variants: 1) When $n$ is odd and $r\in (0, \frac{1}{\sqrt{2}}]$, prove that $|z_1+z_2+\cdots+z_n|\cdot |1/z_1+1/z_2+\cdots+1/z_n| \ge n^2(1-r^2) + r^2$. 2) ($n=3$) Find the minimum of $|z_1+z_2+z_3|\cdot |1/z_1+1/z_2+1/z_3|$.

Can we use z=a+bi?