Korean Mathematical Olympiad Final Round

As $a$, $b$, $c$, $d$ are all odd, we have
$$\begin{aligned} f(n) &\equiv 4n - \left(a\lfloor\frac{n}{a}\rfloor + b\lfloor\frac{n}{b}\rfloor + c\lfloor\frac{n}{c}\rfloor + d\lfloor\frac{n}{d}\rfloor\right) \\ &\equiv \left(n - a\lfloor\frac{n}{a}\rfloor\right) + \left(n - b\lfloor\frac{n}{b}\rfloor\right) + \left(n - c\lfloor\frac{n}{c}\rfloor\right) + \left(n - d\lfloor\frac{n}{d}\rfloor\right) \pmod{2}. \end{aligned}$$
We can observe that $n - a\lfloor\frac{n}{a}\rfloor$ is the remainder of $n$ divided by $a$—denote it as $(n \mod a)$. Meanwhile, the Chinese remainder theorem states that the map
$$\Phi : \{1, \dots, abcd\} \to \{0, \dots, a-1\} \times \{0, \dots, b-1\} \times \{0, \dots, c-1\} \times \{0, \dots, d-1\}, \\ \Phi(n) = (n \mod a, n \mod b, n \mod c, n \mod d)$$
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 $f(n) \equiv x + y + z + w \pmod{2}$, so it follows that
$$\begin{aligned} \sum_{n=1}^{abcd} (-1)^{f(n)} &= \sum_{x=0}^{a-1} \sum_{y=0}^{b-1} \sum_{z=0}^{c-1} \sum_{w=0}^{d-1} (-1)^{x+y+z+w} \\ &= \left(\sum_{x=0}^{a-1} (-1)^x\right) \left(\sum_{y=0}^{b-1} (-1)^y\right) \left(\sum_{z=0}^{c-1} (-1)^z\right) \left(\sum_{w=0}^{d-1} (-1)^w\right) \\ &= 1 \cdot 1 \cdot 1 \cdot 1 = 1. \end{aligned}$$