The total number of seven-digit numbers the sum of

 whose digits is even, is  

Suppose $\begin{array}{l}{x}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}_4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}_5\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}_6\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}_7\end{array}$ represents a seven

digit number. Then x1 takes the value  $1,2,3,\dots ,9$ and

$x_2,x_3,\dots ,x_7$  all take values  $0,1,2,3,\dots ,9$ 

If we keep  $x_1{x}_2,\dots ,x_6$   fixed, then the sum  $x_1+x_2+\dots +x_6$ is

either even or odd. Since x7 takes 10 values $0, 1, 2, ... , 9,$

five of the numbers so formed will be even and 5 odd.   

   Hence, the required number of numbers 

    $=9\cdot 10\cdot 10\cdot 10\cdot 10\cdot 10\cdot 5=4500000$