Two Thousand Fourteen

It is given that:

$\begin{cases} \log{2000xy}-\log{x}\log{y} = 4 &...(1) \\ \log{2yz}-\log{y}\log{z} = 1 &...(2) \\ \log{zx}-\log{z}\log{x} = 0 &...(3) \end{cases} \\ \Rightarrow \begin{cases} \log{2} + 3 + \log{x}+ \log{y}-\log{x}\log{y} = 4 & \\ \Rightarrow \log{x}+ \log{y}-\log{x}\log{y} = 4 - 3 - \log{2} = \log{5} &...(1a) \\ \log{2} + \log{y} + \log{z} -\log{y}\log{z} = 1 & \\ \Rightarrow \log{y} + \log{z} -\log{y}\log{z} = 1 - \log{2} = \log{5} &...(2a) \\ \log{z} + \log{x}-\log{z}\log{x} = 0 & \\ \Rightarrow \log{z} + \log{x}=\log{z}\log{x} &...(3a) \end{cases}$

From Eq. 1a and Eq. 2a: $\Rightarrow x=z$ for all $y$ and this is confirmed by Eq. 3a.

From Eq. 1a, a trivial solution is $x=z=1\space \Rightarrow \log{x} = 0$ and $\space \Rightarrow 0 + \log{y_1} - 0 = \log{5} \space \Rightarrow y_1 = 5$ .

From Eq,3a: $\Rightarrow \log{z} + \log{x}=\log{z}\log{x} \space \Rightarrow 2 \log{x} = (\log{x})^2 \space \Rightarrow \log{x} = 2$

Substituting $\log{x} = 2$ in Eq, 1a:

$\Rightarrow 2 + \log{y_2}-2\log{y_2} = \log{5} \space \Rightarrow \log{y_2} = 2-\log{5} = \log{20}\\ \Rightarrow y_2 = 20$

Therefore, $\space y_1+y_2 = 5 + 20 = \boxed{25}$