jmc

Note that $x_i<1$ for all $i.$

We claim that $$\frac{x}{1-x^2}\ge \frac{3\sqrt{3}}{2}{x}^2$$for all $0<x<1.$ This is equivalent to $2x\ge 3\sqrt{3}{x}^2(1-x^2)=3x^2\sqrt{3}-3x^4\sqrt{3},$ or $$3\sqrt{3}{x}^4-3x^2\sqrt{3}+2x\ge 0.$$We can factor this as $$x(x\sqrt{3}-1{)}^2(x\sqrt{3}+2)\ge 0,$$which clearly holds. Thus, $$\frac{x}{1-x^2}\ge \frac{3\sqrt{3}}{2}{x}^2.$$It follows that $$\frac{x_1}{1-x_1^2}+\frac{x_2}{1-x_2^2}+\frac{x_3}{1-x_3^2}+\cdots +\frac{x_{100}}{1-x_{100}^2}\ge \frac{3\sqrt{3}}{2}(x_1^2+x_2^2+x_3^2+\cdots +x_{100}^2)=\frac{3\sqrt{3}}{2}.$$Equality occurs when $x_1=x_2=x_3=\frac{1}{\sqrt{3}}$ and $x_4=x_5=\cdots =x_{100}=0,$ so the minimum value is $\boxed{\frac{3\sqrt{3}}{2}}.$