jmc

By AM-GM, $$a+b+c+d\ge 4\sqrt[4]{abcd},$$and $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge 4\sqrt[4]{\frac{1}{abcd}},$$so $$(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\ge 4\sqrt[4]{abcd}\cdot 4\sqrt[4]{\frac{1}{abcd}}=16.$$Equality occurs when $a=b=c=d,$ so the minimum value is $\boxed{16}.$