for N,

####...N/2times ####...N/2 times
consider first half N/2 digits

sum(digit root) will always be from 0 to 9
it will be 0 only if all digits are 0's
and it'll will 1,2,3 .... ,9 in other cases...

for N=4
first half will have 2 digits ( total 100 numbers )
1 number(00) will have digit root = 0
99 numbers will have digit root other than 0
( i.e 11 numberes --> digit root 1 , 11 numbers ---> digit root 2 etc )

so for N=4 ,
case i : first half = 00 second half =00 -----> n1 = 1
case ii: first half --> 99 ways , second half --> 11 ways  n2= 99*11

total number = n1+n2 = 99*11 + 1

similarly for N=6 , we have 999*111 + 1
N=8 , 9999*1111 + 1
so on...

Do not use BigInteger to calculate. Let k = n/2, then the result is 10 for k=1; otherwise it is a concatenation of "1"x(k-1), "0", "8"x(k-2), and "90":

1 10
2 1090
3 110890
4 11108890
5 1111088890
6 111110888890
7 11111108888890
8 1111111088888890
9 111111110888888890