Suppose none of the sums of digits are divisible by 11. Let s(x) be the sum of digits of x. Then, s(10x)=s(x). Also, s(10x+1)=s(x)+1, s(10x+2)=s(x)+2, ..., s(10x+9)=s(x)+9. If none of s(x), s(x)+1, ..., s(x)+9 are divisible by 11, then s(x) must be 1 mod 11. Then, s(x)+1=2 mod 11, s(x)+2=3 mod 11, ..., s(x)+9=10 mod 11. The numbers from 10x+10 to 10x+19 are 10(x+1) to 10(x+1)+9. For the same reason as the numbers from 10x to 10x+9, s(x+1)=1 mod 11. The numbers from 10x+20 to 10x+29 are 10(x+2) to 10(x+2)+9. That means that s(x+2)=1 mod 11. Therefore, s(x), s(x+1), and s(x+2) are all 1 mod 11.

Suppose x+1 is divisible by 10. Let x+1=10n. Then, x+2=10n+1. Also, s(x+1)=s(n) and s(x+2)=s(n)+1, so s(x+2)=s(x+1)+1. Since s(x+1)=1 mod 11, s(x+2)=2 mod 11. However, s(x+2)=1 mod 11, so x+1 is not divisible by 10.

Let x+1=10n+m, where m is the last digit of x+1. Since x+1 is not divisible by 10, m>=1. Therefore, x+1>=10n+1, so x>=10n. Since x=10n+(m-1) and x>=10n, m-1 is the last digit of x. Therefore, s(x)=s(n)+(m-1) and s(x+1)=s(n)+m, so s(x+1)=s(x)+1. Since s(x)=1 mod 11, s(x+1)=2 mod 11. However, s(x+1)=1 mod 11, so that is a contradiction. Therefore, at least one of the 39 numbers adds to a multiple of 11.

To find 38 numbers where none of them adds to a multiple of 11, have the tenth number be divisible by 10. Let 10x be that number. Then, 10x+28 is the 38th number, so 10x+29 is not in the set. For the same reasons as above, s(x) and s(x+1) are both 1 mod 11. However, s(x+2) is different. The numbers from 10x+20 to 10x+28 are 10(x+2) to 10(x+2)+8. Then, s(10x+20)=s(x+2), s(10x+21)=s(x+2)+1, ..., s(10x+28)=s(x+2)+8. If none of these are divisible by 11, then s(x+2)=1 or 2 mod 11. We know that 1 mod 11 will not work, so s(x+2)=2 mod 11. If x+1 is not divisible by 10, we get the same contradiction as above. Therefore, x+1 is divisible by 10. When x+1 was divisible by 10 above, we got s(x+2)=2 mod 11. Therefore, this works.

Let x+1=10n. Then, x=10(n-1)+9. Therefore, s(x)=s(n-1)+9 and s(x+1)=s(n). Since s(x)=1 mod 11, s(n-1)=3 mod 11. Since s(x+1)=1 mod 11, s(n)=1 mod 11. For similar reasons as above, if n is not divisible by 10, then s(n-1)+1=s(n). However, s(n-1)+1=4 mod 11 and s(n)=1 mod 11, so n is divisible by 10. Let n=10n'. Then, s(n'-1)=5 mod 11 and s(n')=1 mod 11. s(n'-1)+1=6 mod 11, so n' is divisible by 10. Let n'=10n''. Then, s(n''-1)=7 mod 11 and s(n'')=1 mod 11. s(n''-1)+1=8 mod 11, so n'' is divisible by 10. Let n''=10n'''. Then, s(n'''-1)=9 mod 11 and s(n''')=1 mod 11. s(n'''-1)+1=10 mod 11, so n''' is divisible by 10. Let n'''=10n''''. Then, s(n''''-1)=0 mod 11 and s(n'''')=1 mod 11. s(n''''-1)+1=1 mod 11, so we can stop. The smallest possible value of n'''' is 1. Therefore, x+1=10n=100n'=1000n''=10000n'''=100000n''''=100000. Then, x=99999, so the 10th number in the set is 10x=999990. Then, the first number is 10x-9=999981, and the last number is 10x+28=1000018. Therefore, the numbers from 999981 to 1000018 are 38 consecutive numbers where none adds to a multiple of 11.