jmc

Numbers the rows 1, 2, 3, $\dots ,$ 8 from top to bottom. Let $r_1$ be the row number of the chosen square in the first column. (For example, if the 5th square is chosen in the first column, then $r_1=5.$) Then the label of that square is $\frac{1}{r_1}.$

Similarly, if $r_2$ is the row number of the chosen square in the second column, then its label is $$\frac{1}{r_2+1}.$$In general, let $r_i$ be the row number of the chosen square in column $i,$ so its label is $$\frac{1}{r_i+i-1}.$$Then we want to minimize $$\frac{1}{r_1}+\frac{1}{r_2+1}+\frac{1}{r_3+2}+\cdots +\frac{1}{r_8+7}.$$By AM-HM, $$\frac{r_1+(r_2+1)+(r_3+2)+\cdots +(r_8+7)}{8}\ge \frac{8}{\frac{1}{r_1}+\frac{1}{r_2+1}+\frac{1}{r_3+2}+\cdots +\frac{1}{r_8+7}},$$so $$\begin{array}{rl}\frac{1}{r_1}+\frac{1}{r_2+1}+\frac{1}{r_3+2}+\cdots +\frac{1}{r_8+7}& \ge \frac{64}{r_1+(r_2+1)+(r_3+2)+\cdots +(r_8+7)}\\ & =\frac{64}{r_1+r_2+r_3+\cdots +r_8+28}.\end{array}$$Since there exists one chosen square in each row, $r_1,$ $r_2,$ $r_3,$ $\dots ,$ $r_8$ are equal to 1, 2, 3, $\dots ,$ 8 in some order. Therefore, $$\frac{1}{r_1}+\frac{1}{r_2+1}+\frac{1}{r_3+2}+\cdots +\frac{1}{r_8+7}\ge \frac{64}{1+2+3+\cdots +8+28}=\frac{64}{36+28}=1.$$Equality occurs when we choose all eight squares labelled $\frac{1}{8},$ so the smallest possible sum is $\boxed{1}.$