Find the sum of all positive solutions of \[2\cos(2x)\left(\cos(2x)-\cos\left(\frac{2020\pi^2}x\right)\right)=\cos(4x)-1.\]

1. Using $\cos4x-1=2\cos^22x-2$, the equation reduces to $\cos2x\cos(K/x)=1$ where $K=2020\pi^2$, forcing both cosines to be $+1$ or both $-1$.
2. The ``both $-1$'' case is impossible (would require a product of two odd numbers to be even). The ``both $+1$'' case gives $x=m\pi$ where $m$ is a positive divisor of $1010=2\cdot5\cdot101$.
3. Divisors of $1010$: $1,2,5,10,101,202,505,1010$, summing to $1836$. The answer is $\boxed{1836\pi}$.