This solution looks kind of straightforward, if anything looks fishy please let me know. Let $a_1<a_2<\dots<a_{13}$ be our $3-$digit $13$ positive inntegers. Then $\frac{a_{13}}{a_1}<\frac{1000}{100}=10$. We may write $$10>\frac{a_{13}}{a_1}=\prod_{i=1}^{12}\frac{a_{i+1}}{a_i},$$therefore there exists some $i$ such that $\frac{a_{i+1}}{a_i}<\sqrt[12]{10}$. Next, removing from this product the factor $\frac{a_{i+1}}{a_i}$ and also the fractions $\frac{a_{i+2}}{a_{i+1}}$ and $\frac{a_i}{a_{i-1}}$ (for $i=1$ or $i=12$, remove only the one that actually exists), we get $$10>\prod_{\substack{1\le j\le 12\\ j\ne i-1,i,i+1}}\frac{a_{j+1}}{a_j},$$so for some $j\ne i-1,i,i+1$, $\frac{a_{j+1}}{a_j}<\sqrt[9]{10}$ holds. In a similar manner we obtain some $k\ne j-1,j,j+1,i-1,i,i+1$ such that $\frac{a_{k+1}}{a_k}<\sqrt[6]{10}$. It follows that $k,k+1,j,j+1,i,i+1$ are pairwise distinct and $$3=1+1+1<\frac{a_{i+1}}{a_i}+\frac{a_{j+1}}{a_j}+\frac{a_{k+1}}{a_k}<\sqrt[12]{10}+\sqrt[9]{10}+\sqrt[6]{10}<3,98<4.$$By taking the set $S=\{a_i,a_{i+1},a_j,a_{j+1},a_k,a_{k+1}\}$, we are done.