The 37th Korean Mathematical Olympiad Final Round

As $a$, $b$, $c$, $d$ are all odd, we have $$\begin{array}{rl}f(n)& \equiv 4n-\left(a\lfloor \frac{n}{a}\rfloor +b\lfloor \frac{n}{b}\rfloor +c\lfloor \frac{n}{c}\rfloor +d\lfloor \frac{n}{d}\rfloor \right)\\ & \equiv \left(n-a\lfloor \frac{n}{a}\rfloor \right)+\left(n-b\lfloor \frac{n}{b}\rfloor \right)+\left(n-c\lfloor \frac{n}{c}\rfloor \right)+\left(n-d\lfloor \frac{n}{d}\rfloor \right)(\mathrm{mod}2).\end{array}$$ We can observe that $n-a\lfloor \frac{n}{a}\rfloor$ is the remainder of $n$ divided by $a$—denote it as $(n\mathrm{mod}a)$. Meanwhile, the Chinese remainder theorem states that the map $$\begin{gathered} \Phi :\{1,\dots ,abcd\}\to \{0,\dots ,a-1\}\times \{0,\dots ,b-1\}\times \{0,\dots ,c-1\}\times \{0,\dots ,d-1\}, \\ \Phi (n)=(n\mathrm{mod}a,n\mathrm{mod}b,n\mathrm{mod}c,n\mathrm{mod}d) \end{gathered}$$ is a one-to-one correspondence, so we can replace our sum in $n$ as a sum in $(x,y,z,w)=\Phi (n)$ when $(x,y,z,w)$ ranges over all elements of $\{0,\dots ,a-1\}\times \{0,\dots ,b-1\}\times \{0,\dots ,c-1\}\times \{0,\dots ,d-1\}$. If $\Phi (n)=(x,y,z,w)$ then $f(n)\equiv x+y+z+w(\mathrm{mod}2)$, so it follows that $$\begin{array}{rl}\sum _{n=1}^{abcd}(-1{)}^{f(n)}& =\sum _{x=0}^{a-1}\sum _{y=0}^{b-1}\sum _{z=0}^{c-1}\sum _{w=0}^{d-1}(-1{)}^{x+y+z+w}\\ & =\left(\sum _{x=0}^{a-1}(-1{)}^x\right)\left(\sum _{y=0}^{b-1}(-1{)}^y\right)\left(\sum _{z=0}^{c-1}(-1{)}^z\right)\left(\sum _{w=0}^{d-1}(-1{)}^w\right)\\ & =1\cdot 1\cdot 1\cdot 1=1.\end{array}$$