jmc

By Cauchy-Schwarz, $$(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\right)\ge (1+1+2+4{)}^2=64.$$Equality occurs when $a=b=\frac{c}{2}=\frac{d}{4}$ and $a+b+c+d=1.$ We can solve to get $a=\frac{1}{8},$ $b=\frac{1}{8},$ $c=\frac{1}{4},$ and $d=\frac{1}{2},$ so the minimum value is $\boxed{64}.$