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Solution:

∵ 19X？ 2X+2Y+Z,∴ 17X-2Y？ Z ( 1)

∵4Y？ 2X+2Y+Z,∴2Y-2X？ Z (2)

z？ 2X+2Y ∴2X +2Y？ Z (3)

Let (1) be 17x-2y < z (4).

Then 19X? 2X+2Y+Z is 19x < 2x+2y+z,

Let (2) be 2y-2x < z (5)

And then 4Y? 2X+2Y+Z，4y < 2x+2y+z，

(4)+(5)， 15x < 2z (6)

Let (3) be 2X+2Y≤Z(7)

Then z? 2X+2Y is Z≧2X+2Y.

(4)+(7)， 19x < 2z (8)

From [(5)+(7)]/2, 2y < z (9)

X, y and z must all be greater than 0 and must be an integer multiple of 100.

According to (8) formula (9) and (4) formula 17x-2y < z,

When Z≥ 1000, X= 100, Y=400.

P.S.: "X, Y and Z must all be greater than 0 and must be an integer multiple of 100" is the key to solve this problem. I didn't pay enough attention to this for a long time, which puzzled me all day and almost missed the solution of this problem. I suddenly realized this key point this morning, so the problem was solved.