74th Romanian Mathematical Olympiad

a) If $d_1,d_2,\dots ,d_p$ are the positive divisors of a positive integer $n$, then $\{d_1,d_2,\dots ,d_p\}=\{\frac{n}{d_1},\frac{n}{d_2},\dots ,\frac{n}{d_p}\}$. Let $b_1,b_2,\dots ,b_q$ be the positive divisors of $b$. Then $$Q(a,b)\le \frac{a}{b_1}+\cdots +\frac{a}{b_q}=\frac{a}{b}\left(\frac{b}{b_1}+\cdots +\frac{b}{b_q}\right)=\frac{a}{b}S(b)=\frac{a}{b}Q(b,a),(1)$$ therefore $\frac{Q(a,b)}{a}\le \frac{Q(b,a)}{b}$.

Since the statement is symmetrical with respect to $a$ and $b$, $\frac{Q(b,a)}{b}\le \frac{Q(a,b)}{a}$, hence $\frac{Q(b,a)}{b}=\frac{Q(a,b)}{a}$, which shows that (1) is an equality. This shows that $a$ is divisible with all the divisors of $b$ and $b$ is divisible with all the divisors of $a$, so $a=b$.

b) It is not true. For instance, if $a=2$ and $b=5$, then $S(2)+S(5)=(1+2)+(1+5)=9$ and $Q(2,5)+Q(5,2)=2+(5+2)=9$, but $a\ne b$.