Could it be this is not from Putnam 1980?

Peter wrote: Could it be this is not from Putnam 1980? Yeah, it is not from Putnam 1980. It seems that some problems in G section has wrong references. I'm gonna check it.

any solution？

This is known as Hurwitz's Theorem. Note that the following beautiful proof is not mine. It's enough to consider only $\xi \in (0,1)$. For each positive integer $n$, consider the Farey sequence $F_n$. Suppose $\frac{a}{b},\frac{c}{d}$ are two consecutive terms in $F_n$ that $\frac{a}{b}<\xi <\frac{c}{d}$. We'll show that at least one of $\frac{a}{b},\frac{c}{d},\frac{a+c}{b+d}$ work. Note that this complete the problem by inductively choosing large $n$. Suppose to the contrary that none of three fractions work. We get that $\xi -\frac{a}{b}\geqslant \frac{1}{\sqrt{5}b^2}$ and $\frac{c}{d} -\xi \geqslant \frac{1}{\sqrt{5}d^2}$. Adding two inequalities and using $bc-ad=1$ gives us $\frac{1}{bd}\geqslant \frac{1}{\sqrt{5}b^2}+\frac{1}{\sqrt{5}d^2}\implies bd\geqslant \frac{b^2+d^2}{\sqrt{5}}$. Now assume $\xi >\frac{a+c}{b+d}$ (the other case can be done in similar way.) We get $\xi -\frac{a+c}{b+d} \geqslant \frac{1}{\sqrt{5}(b+d)^2}$. So, $\frac{1}{d(b+d)}=\frac{c}{d}-\frac{a+c}{b+d}\geqslant \frac{1}{\sqrt{5}d^2}+\frac{1}{\sqrt{5}(b+d)^2}\implies d(b+d)\geqslant \frac{d^2+(b+d)^2}{\sqrt{5}}$. Hence, $d^2+2bd\geq \frac{b^2+2d^2+(b+d)^2}{\sqrt{5}}\implies -\frac{((\sqrt{5}-1)d-2b)^2}{2\sqrt{5}} \geqslant 0$, which is impossible, done.

Actually it's sufficient to consider $n$ only as a prime number, because then none of fractions $\frac{a}{b},\frac{c}{d}$ repeat and we know there are infinitely many prime numbers.