Answer: 25

Solution:

Let $a = \log_{10} x,$ $b = \log_{10} y,$ $c = \log_{10} z.$ Using $\log_{10} 2000 = 4 - \log_{10} 5,$ the first equation becomes $a + b - ab = \log_{10} 5,$ which factors as $(1 - a)(1 - b)$ $= 1 - \log_{10} 5$ $= \log_{10} 2.$ Similarly the second equation gives $(1 - b)(1 - c)$ $= 1 - (1 - \log_{10} 2)$ $= \log_{10} 2,$ and the third gives $(1 - c)(1 - a) = 1.$

Dividing the first two (note $1 - b \ne 0$) yields $1 - a = 1 - c,$ and then the third equation gives $(1 - a)^2 = 1,$ so $1 - a = \pm 1.$ If $1 - a = 1,$ then $1 - b = \log_{10} 2,$ so $b = 1 - \log_{10} 2 = \log_{10} 5$ and $y = 5$ (indeed $(x, y, z) = (1, 5, 1)$ works). If $1 - a = -1,$ then $1 - b = -\log_{10} 2,$ so $b = \log_{10} 20$ and $y = 20$ (from $(x, y, z) = (100, 20, 100)$).

Therefore $y_1 + y_2 = 5 + 20 = 25.$