$\underline{\textbf{Solution 1}}$ One way to prove this is to use the Cauchy inequality. (see # 3865.) Let $b_i=2-a_i$, where $i=1, 2, 3,\cdots, n$. Then we have $b_1+b_2+\cdots + b_n = 2n-1$. It follows that $$ \begin{array}{lrcl} & \frac{a_1}{2-a_1} + \frac{a_2}{2-a_2}+\cdots+\frac{a_n}{2-a_n} &\ge&\frac{n}{2n-1}\\ \Leftrightarrow & \frac{2-b_1}{b_1} + \frac{2-b_2}{b_2} + \cdots + \frac{2-b_n}{b_n} &\ge&\frac{n}{2n-1}\\ \Leftrightarrow & \frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_n} &\ge&\frac{n^2}{2n-1}\\ \Leftrightarrow & \frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_n} &\ge&\frac{n^2}{b_1+b_2+\cdots + b_n}\\ \Leftrightarrow & \Big(\frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_n}\Big)(b_1+b_2+\cdots + b_n) &\ge& n^2 \end{array} $$ The last relation obviously holds by the Cauchy inequality.

$\underline{\textbf{Solution 2}}$ This can also be proved by utilizing the sum of infinite geometric sequences reversely. Because $0 < a_1 < 1$, it must have $|\frac{a_1}{2}|<1$. Therefore $$\frac{a_1}{2-a_1}=\frac{\frac{a_1}{2}}{1-\frac{a_1}{2}}=\frac{a_1}{2}+\big(\frac{a_1}{2}\big)^2+\big(\frac{a_1}{2}\big)^3+\cdots+\big(\frac{a_1}{2}\big)^n+\cdots$$ It follows that \begin{align*} &\frac{a_1}{2-a_1} + \frac{a_2}{2-a_2}+\cdots+\frac{a_n}{2-a_n}\ge\frac{n}{2n-1}\\ =&\Big(\frac{a_1}{2}+\frac{a_2}{2}+\cdots + \frac{a_n}{2}\Big) +\\ &\Big(\big(\frac{a_1}{2}\big)^2+\big(\frac{a_2}{2}\big)^2+\cdots + \big(\frac{a_n}{2}\big)^2\Big)+\\ &\cdots\\ &\Big(\big(\frac{a_1}{2}\big)^n+\big(\frac{a_2}{2}\big)^n+\cdots + \big(\frac{a_n}{2}\big)^n\Big)+\\ &\cdots\\ \ge&\frac{1}{2}+\frac{\big(\frac{a_1}{2}+\frac{a_2}{2}+\cdots+\frac{a_n}{2}\big)^2}{n}+\frac{\big(\frac{a_1 }{2}+\frac{a_2}{2}+\cdots+\frac{a_n}{2}\big)^3}{n^2}\\ &+\cdots+\frac{\big(\frac{a_1}{2}+\frac{a_2}{2}+\cdots+\frac{a_n}{2}\big)^n}{n^{n-1}} +\cdots\\ =&\frac{1}{2}+\frac{1}{2^2n}+\frac{1}{2^3n^2}+\cdots +\frac{1}{2^nn^{n-1}}+\cdots\\ =&\frac{\frac{1}{2}}{1-\frac{1}{2n}}=\frac{n}{2n-1} \end{align*}