4. Given real numbers $a, b, c, d$ satisfy: for any real number $x$,
$a \cos x + b \cos 2x + c \cos 3x + d \cos 4x \leq 1$.
Find the maximum value of $a + b - c + d$ and the values of the real numbers $a, b, c, d$ at that time.
(Supplied by Li Shenghong)
4. Let $f(x)=a \cos x+b \cos 2 x+$ $c \cos 3 x+d \cos 4 x$.

From $f(0)=a+b+c+d$,
$f(\pi)=-a+b-c+d$,
$f\left(\frac{\pi}{3}\right)=\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2}$,
then $a+b-c+d$
$$
=f(0)+\frac{2}{3} f(\pi)+\frac{4}{3} f\left(\frac{\pi}{3}\right) \leqslant 3 .
$$

Equality holds if and only if $f(0)=f(\pi)=f\left(\frac{\pi}{3}\right)=1$, i.e.,
$$
a=1, b+d=1, c=-1
$$

In this case, let $t=\cos x(-1 \leqslant t \leqslant 1)$. Then
$$
\begin{array}{l} 
f(x)-1 \\
= \cos x+b \cos 2 x-\cos 3 x+d \cos 4 x-1 \\
= t+(1-d)\left(2 t^{2}-1\right)-\left(4 t^{3}-3 t\right)+ \\
d\left(8 t^{4}-8 t^{2}+1\right)-1 \\
= 2(t-1)(t+1)(2 t-1)[2 d t-(1-d)] \\
\leqslant 0
\end{array}
$$

for any real number $t \in[-1,1]$.
Thus, $d>0$, and $\frac{2 d}{2}=\frac{1-d}{1}$, i.e., $d=\frac{1}{2}$.
Therefore, the maximum value of $a+b-c+d$ is 3, and at this time
$$
(a, b, c, d)=\left(1, \frac{1}{2},-1, \frac{1}{2}\right) \text {. }
$$