As Andrei Lazanu from Tudor Vianu National College, Bucharest, Romania says "The only solution for the problem is $a=-c$ and $b=d$ so that the sum

${\displaystyle \frac{a}{b}+\frac{c}{d}}$

is zero and

${\displaystyle \frac{a+c}{b+d}=0,}$

so the relation is true".

Why is this the only possibility? The relation

${\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}}$

holds if and only if

${\displaystyle \frac{\mathrm{ad}+\mathrm{bc}}{\mathrm{bd}}=\frac{a+c}{b+d}}$

that is

${\displaystyle (\mathrm{ad}+\mathrm{bc})(b+d)=(a+c)\mathrm{bd}}$

which holds if and only if

${\displaystyle {\mathrm{ad}}^2+b^{2}c=0,}$

or equivalently

${\displaystyle {\mathrm{ad}}^2=-b^{2}c.}$

We know that any whole number can be written as the unique product of prime factors. As $a$ and $b$ are coprime and $c$ and $d$ are coprime ${\mathrm{ad}}^2=-b^{2}c$ is true only if $a$ divides $c$ and $c$ divides $a$ and they are of opposite sign, that is $a=-c$. Thus the original formula holds if and only if $d=\pm b$ and $b+d\ne 0$ that is $b=d$ so the formula holds if and only if $\frac{a}{b}=\frac{-c}{d}$ and their sum is zero.