16. If any group of numbers is taken from $1,2, \cdots, 50$, there will always be two different numbers whose product is an integer multiple of their sum. How many numbers must be taken at minimum?
Consider $(a+b) \mid a b$, let $(a, b)=d \Rightarrow a=a_{1} d, b=b_{1} d,\left(a_{1}, b_{1}\right)=1$.
Then $\left(a_{1}+b_{1}\right)\left|a_{1} b_{1} d \Rightarrow\left(a_{1}+b_{1}\right)\right| d \Rightarrow a_{1}+b_{1} \leqslant d \Rightarrow\left(a_{1}+b_{1}\right)^{2} \leqslant\left(a_{1}+b_{1}\right) d$ $=a+b \leqslant 99 \Rightarrow 3 \leqslant a_{1}+b_{1} \leqslant 9,3 \leqslant d \leqslant 33$. Without loss of generality, assume $a_{1}<b_{1}$, $a_{1}+b_{1}=3$, at this time $\left(a_{1}, b_{1}\right)=(1,2), 3 \mid d, \quad(a, b)=(3,6),(6,12),(9,18),(12,24)$, $(15,30),(18,36),(21,42),(24,48)$;
$a_{1}+b_{1}=4$, at this time $\left(a_{1}, b_{1}\right)=(1,3), 4 \mid d,(a, b)=(4,12),(8,24),(12,36),(16,48)$; $a_{1}+b_{1}=5$, at this time $\left(a_{1}, b_{1}\right)=(1,4),(2,3), 5 \mid d,(a, b)=(5,20),(10,40),(10,15)$, $(20,30),(30,45)$;
$a_{1}+b_{1}=6$, at this time $\left(a_{1}, b_{1}\right)=(1,5), 6 \mid d,(a, b)=(6,30)$;
$a_{1}+b_{1}=7$, at this time $\left(a_{1}, b_{1}\right)=(1,6),(2,5),(3,4), 7 \mid d,(a, b)=(7,42),(14,35)$, $(21,28)$;
$a_{1}+b_{1}=8$, at this time $\left(a_{1}, b_{1}\right)=(1,7),(3,5), 8 \mid d,(a, b)=(24,40)$; $a_{1}+b_{1}=9$, at this time $\left(a_{1}, b_{1}\right)=(1,8),(2,7),(4,5), 9 \mid d,(a, b)=(36,45)$.
The above 23 pairs $(a, b)$ satisfy $a, b \in S$ and $(a+b) \mid a b$. Take the smallest $M=\{12,24,30,10,42$, $18,6,48,20,35,28,45\}$ (not unique) containing at least one number from each of the above 23 pairs, then $S \backslash M$ does not contain two numbers that satisfy the condition, so at least 39 numbers need to be taken. On the other hand, 12 pairs $(3,6),(16,48),(5,20),(14,35),(21,28),(36,45),(9,18),(4,12),(8,24),(15,30)$, $(10,40),(7,42)$ are mutually exclusive, for any 39-element subset $R$ of $S$, the 11 removed numbers belong to at most 11 of the 12 pairs, i.e., there must be one pair in $R$. In summary, at least 39 numbers need to be taken.