Argentina_2018

Player B has a winning strategy. One way to see this is to divide the 3-digit numbers into 8 groups with the following property: For every two numbers $a$, $b$ from the same group, $a>b$, one has $\frac{a}{b}<\frac{4}{3}$.

$$\begin{array}{rl}{G}_1& =\{100,\dots ,133\},\\ G_2& =\{134,\dots ,178\},\\ G_3& =\{179,\dots ,238\},\\ G_4& =\{239,\dots ,318\},\\ G_5& =\{319,\dots ,425\},\\ G_6& =\{426,\dots ,567\},\\ G_7& =\{568,\dots ,757\},\\ G_8& =\{758,\dots ,999\}.\end{array}$$

For a justification it suffices to note that the ratio of the last number and the first number in a group is less than $\frac{4}{3}$, for instance, $\frac{567}{426}<\frac{4}{3}$.

Since there are 13 different numbers chosen by A and $13>8$, some two of them are in the same group $G_i$, $1<i<8$. Let them be $a$ and $b$ with $a>b$; then $1<\frac{a}{b}<\frac{4}{3}$. Among the remaining 11 numbers B can find another two with the same property, say $c$ and $d$ with $1<\frac{c}{d}<\frac{4}{3}$. This is because $11>8$. Finally B can select one more analogous pair $e,f$, satisfying $1<\frac{e}{f}<\frac{4}{3}$, because there are still $9>8$ numbers left. Now adding up gives $3=1+1+1<\frac{a}{b}+\frac{c}{d}+\frac{e}{f}<3\cdot \frac{4}{3}=4$, and the task of B is complete.