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Consider the numbers $2000,2002,\cdots,2998$ Count of these numbers $=\dfrac{2998-2000}{2}+1=500$

Similarly, each of $(4000,\cdots,4998),$ $(6000,\cdots,6998),$ $(8000,\cdots,8998)$ gives $500$ numbers.

So, required count $500+500+500+500=2000$

Solution 2 (using combinatorics)

First digit can be any of $(2,4,6,8)$ Second digit can be any of the $10$ digits Third digit can be any of the $10$ digits Fourth digit can be any of $(0,2,4,6,8)$