China Southeastern Mathematical Olympiad

Since
$$\begin{aligned} f(0) &= a + b + c + d, \\ f(\pi) &= -a + b - c + d, \\ f\left(\frac{\pi}{3}\right) &= \frac{a}{2} - \frac{b}{2} - c - \frac{d}{2}, \end{aligned}$$
then
$$a + b - c + d = f(0) + \frac{2}{3}f(\pi) + \frac{4}{3}f\left(\frac{\pi}{3}\right) \le 3$$
if and only if $f(0) = f(\pi) = f(\frac{\pi}{3}) = 1$, that is, if $a = 1$, $b + d = 1$ and $c = -1$, then the equality holds. Let $t = \cos x$, $-1 \le t \le 1$. Then
$$\begin{aligned} f(x) - 1 &= \cos x + b \cos 2x - \cos 3x + d \cos 4x - 1 \\ &= t + (1-d)(2t^2 - 1) - (4t^3 - 3t) + d(8t^4 - 8t^2 + 1) - 1 \\ &= 2(1 - t^2)[-4dt^2 + 2t + (d-1)] \le 0, \forall t \in [-1, 1], \end{aligned}$$
that is
$$4dt^2 - 2t + (1-d) \ge 0, \forall t \in (-1, 1).$$
Taking $t = 1/2 + \epsilon$, $|\epsilon| < 1/2$, then $\epsilon[(2d - 1) + 4d\epsilon] \ge 0$, $|\epsilon| < 1/2$. So we see that $d = \frac{1}{2}$. If $d = \frac{1}{2}$, then
$$4dt^2 - 2t + (1-d) = 2t^2 - 2t + 1/2 = 2(t - 1/2)^2 \ge 0.$$
So, the maximal number of $a+b-c+d$ is $3$, and $(a, b, c, d) = (1, \frac{1}{2}, -1, \frac{1}{2})$. $\square$