jmc

By Cauchy-Schwarz, $$(a_1+a_2+\cdots +a_{12})\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_{12}}\right)\ge (1+1+\cdots +1{)}^2=12^2=144,$$so $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_{12}}\ge 144.$$Equality occurs when $a_i=\frac{1}{12}$ for all $i,$ so the minimum value is $\boxed{144}.$