You can bound it, but can you construct it?

The set $S = \{ 17,19,23,25,27,29,31,32,37,41,43,47,49\}$ shows that it is possible to find a set of $13$ mutually coprime positive integers with range $32$ .

Suppose that it were possible to find a set of $13$ mutually coprime positive integers with range less than $32$ . At most one of these integers can be even, and so we would have to be able to find a set $T$ of $12$ mutually coprime odd positive integers with range less than $32$ . This set $T$ would be a subset of $T_0 = \{n,n+2,n+4,n+6,n+8,n+10,n+12,n+14,n+16,n+18,n+20,n+22,n+24,n+26,n+28,n+30\}$ for some odd $n \in T$ . Note that $T_0$ has $16$ elements. Now all the members of one of the three sets $\{n,n+6,n+12,n+18,n+24,n+30\} \hspace{1cm} \{n+2,n+8,n+14,n+20,n+26\} \hspace{1cm} \{n+4,n+10,n+16,n+22,n+28\}$ are multiples of $3$ . Whichever of the three sets it is, $T$ can only contain one element from that set, and so $4$ or $5$ elements of $T_0$ will not belong to $T$ . Since $T$ is meant to contain $12$ elements, we must have only discarded $4$ elements from $T_0$ . Thus we deduce that $n$ is not a multiple of $3$ , and hence we one of the two sets $A = \{n+2,n+8,n+14,n+20,n+26\} \hspace{1cm} B = \{n+4,n+10,n+16,n+22,n+28\}$ consists of multiples of $3$ , and we must discard four elements of the relevant set ( $A$ or $B$ ) to ensure that no two elements of $T$ are multiples of $3$ . Having discarded these four elements, we have obtained the set $T$ . Now consider the sets $\{n,n+10,n+20,n+30\} \hspace{1cm} \{n+2,n+12,n+22\} \hspace{1cm} \{n+4,n+14,n+24\} \hspace{1cm} \{n+6,n+16,n+26\} \hspace{1cm} \{n+8,n+18,n+28\}$ No two elements of $A$ differ by $10$ , and so no two elements of $A$ belong to the same one these five sets. Similarly no two elements of $B$ belong to the same one of these five sets. Thus, whichever four elements have been removed from $T_0$ , and whichever of the above five sets contains the multiples of $5$ , we see that $T$ will contain two multiples of $5$ , which is not possible.

Thus the least possible range is $\boxed{32}$ .