Let m be a natural number such that 2000 < m < 6000 and let k be the sum of all the digits in m. Then the numbers for which k is  even  is
 

$\mathrm{consider} 5 \mathrm{digited} \mathrm{number}$ $abcde$

$\mathrm{first} \mathrm{place} a \mathrm{can} \mathrm{be} 2,3,4,5 \mathrm{in} 4 \mathrm{ways}$

$\mathrm{b},\mathrm{c},\mathrm{d} \mathrm{can} \mathrm{be} \mathrm{any} \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{digits} \mathrm{from} 0 \mathrm{to} 9$

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{digits} \mathrm{is} \mathrm{even} \mathrm{f} \mathrm{can} \mathrm{be} \mathrm{filled} \mathrm{in} 5 \mathrm{ways} \\ (\mathrm{f}=\mathrm{odd} \mathrm{if} \mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d} \mathrm{is} \mathrm{odd} \mathrm{and} \mathrm{f}= \mathrm{even} \mathrm{if} \mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d} \mathrm{is} \mathrm{even}) \end{gathered}$$

$$\begin{gathered} total number of 5 digit numbers whose sum of digits \\ is even and lie between 2000 and 6000 \end{gathered}$$

$Required number=4\left({10}^3\right)5-1=20000-1=19999$