25. The equation
$$
2 \cos 2 x \cdot\left(\cos 2 x-\cos \frac{2014 \pi^{2}}{x}\right)=\cos 4 x-1
$$

has the sum of all its positive real roots as ( ).
(A) $\pi$
(B) $810 \pi$
(C) $1008 \pi$
(D) $1080 \pi$
(E) $1800 \pi$
25. D.

From the problem, we have
$2 \cos 2 x \cdot\left(\cos 2 x-\cos \frac{2014 \pi^{2}}{x}\right)=2 \cos ^{2} 2 x-2$.
Let $a=\cos 2 x, b=\cos \frac{2014 \pi^{2}}{x}$. Then
$|a| \leqslant 1,|b| \leqslant 1$.
Thus, the original equation can be transformed into
$$
2 a^{2}-2 a b=2 a^{2}-2 \Rightarrow a b=1 \text {. }
$$

Since $|a| \leqslant 1,|b| \leqslant 1$, we have
$a=b=1$ or $a=b=-1$.
When $a=b=1$,
$\cos 2 x=\cos \frac{2014 \pi^{2}}{x}=1$
$\Rightarrow\left\{\begin{array}{l}x=k_{1} \pi, \\ x=\frac{1007 \pi}{k_{2}}\end{array} k_{1}, k_{2} \in \mathbf{Z}_{+}\right)$.
Thus, $x=\pi, 19 \pi, 53 \pi, 1007 \pi$.
When $a=b=-1$,
$\cos 2 x=\cos \frac{2014 \pi^{2}}{x}=-1$
$\Rightarrow\left\{\begin{array}{l}x=\frac{\left(2 k_{3}+1\right) \pi}{2}, \\ x=\frac{4028 \pi}{2 k_{4}+1}\end{array} \quad\left(k_{3} 、 k_{4} \in \mathbf{Z}_{+}\right)\right.$.
If $\frac{\left(2 k_{3}+1\right) \pi}{2}=\frac{4028 \pi}{2 k_{4}+1}$, then
$\left(2 k_{3}+1\right)\left(2 k_{4}+1\right)=8056$.
Since the left side of the above equation is odd and the right side is even, there is no solution.
In summary, the sum of all positive real roots of the equation is
$$
\pi+19 \pi+53 \pi+1007 \pi=1080 \pi \text {. }
$$