jmc

By Cauchy-Schwarz, $$(a+b+c+d+e+f)\left(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}+\frac{16}{d}+\frac{25}{e}+\frac{36}{f}\right)\ge (1+2+3+4+5+6{)}^2=441,$$so $$\frac{1}{a}+\frac{4}{b}+\frac{9}{c}+\frac{16}{d}+\frac{25}{e}+\frac{36}{f}\ge \frac{441}{7}=63.$$Equality occurs when $a^2=\frac{b^2}{4}=\frac{c^2}{9}=\frac{d^2}{16}=\frac{e^2}{25}=\frac{f^2}{36}$ and $a+b+c+d+e+f=7.$ Solving, we find $a=\frac{1}{3},$ $b=\frac{2}{3},$ $c=1,$ $d=\frac{4}{3},$ $e=\frac{5}{3},$ and $f=2,$ so the minimum value is $\boxed{63}.$