We know $\displaystyle\sum_{i=1}^{13} a_{2i-1} = \displaystyle\sum_{i=1}^{13} a_{2i} = c$ for some constant $1 \le c \le 13$ . For each case of $c$ , we must have $a_1 = 1$ , so we can set $a_3, a_5, \cdots , a_{25}$ in $\dbinom{12}{c-1}$ ways, and we can set $a_2, a_4, \cdots , a_{26}$ in $\dbinom{13}{c}$ ways

Therefore our answer is $\sum_{i=1}^{13} \dbinom{12}{c-1}\dbinom{13}{c} = \sum_{i=1}^{13} \dbinom{12}{13-c}\dbinom{13}{c} = \dbinom{25}{13} \equiv \boxed{300} \pmod{1000}$ by Vandermonde's Identity.

Note that $a_1 = 1$ , or else it would not be a $26$ digit number. The number of ways to have the LHS digit sum be equal to $n$ and the RHS digit sum equal to $n$ is equal to ${12 \choose {n-1}}$ and ${13 \choose n}$ , respectively. We could find the sum from $n = 1 to 13$ of that product, but I found an easier way through some investigation:

Note that the sum is taking two rows in pascal's triangle and taking the element in the top row and multiplying it with the number that is to the bottom and to the right of it. I found the first sums and found a sequence of $1, 3, 10, 35, 126$ , which I recognized as binomial coefficients, so I converted them: ${1 \choose 0}, {3 \choose 2}, {5 \choose 3}, {7 \choose 4}, {9 \choose 5}$ , and I found a pattern that it was equal to ${{2n+1} \choose {n+1}}$ , though due to the fact that I should really be doing college apps I left it at that, computed ${{25} \choose {13}} \pmod {1000}$ by finding the last digit and multiplying by $100$ and called it a day.

We proceed with casework on the possible sums. If the sum is 1, then there is a total of $\binom{12}{0}\cdot\binom{13}{1}$ (there must be a 1 as the first digit and the digits at the odd position have $\binom{12}{0}$ and the digits at the even position have $\binom{13}{1}$ ). Similarly, if the sum is 2, then there is a total of $\binom{12}{1} \cdot \binom{13}{2}$ . Continuing this gives us the sum $\binom{12}{0} \cdot \binom{13}{1} + \binom{12}{1} \cdot \binom{13}{2} + \cdots + \binom{12}{12} \cdot \binom{13}{13} .$ This is just simply Vandermonde's identity, so we have $\binom{25}{13}$ which has last three digits $\boxed{300}$ .

Sorry for sharing this here..... i am a beginner thats d reason i took this method :( First digit in 26-digit number should be 1.. so for remaining 25 digits 1==even digits-odd digits and no of ways ==5200\boxed{300}

here its c programme

include<stdio.h>

int main(){ int q,w,e,r,t,y,u,i,o,p,a,s,d,f,g,h,j,k,l,z,x,c,v,b,n,sum=0; for(q=0;q<2;q++){
for(w=0;w<2;w++){ for(e=0;e<2;e++){ for(r=0;r<2;r++){ for(t=0;t<2;t++){ for(y=0;y<2;y++){ for(u=0;u<2;u++){ for(i=0;i<2;i++){ for(o=0;o<2;o++){ for(p=0;p<2;p++){ for(a=0;a<2;a++){ for(s=0;s<2;s++){ for(d=0;d<2;d++){ for(f=0;f<2;f++){ for(g=0;g<2;g++){ for(h=0;h<2;h++){ for(j=0;j<2;j++){ for(k=0;k<2;k++){ for(l=0;l<2;l++){ for(z=0;z<2;z++){ for(x=0;x<2;x++){ for(c=0;c<2;c++){ for(v=0;v<2;v++){ for(b=0;b<2;b++){ for(n=0;n<2;n++){ if(q+w+e+r+t+y+u+i+o+p+a+s+d-f-g-h-j-k-l-z-x-c-v-b-n==1) sum++; }}}}}}}}}}}}}}}}}}}}}}}}} printf("%d", sum); return 0; }