1Suppose that the smallest number in your sequence of 39 consecutive natural numbers is n. Let k be the smallest number in your sequence that has a zero for the units digit. k is at most n+9. Your sequence then has three blocks of ten numbers each k,k+1,...,k+9; k+10,k+11,...,k+19; and k+20,k+21,...,k+29. The first number in each block (k, k+10 and k+20) has a zero for the units digit. Let m be the smallest of these three numbers that does not have a nine for the tens digit. m is either k or k+10. There are thus 20 consecutive numbers in your sequence, m,m+1,...,m+19 where the units digit of m is zero and the tens digit of m is not 9.
Let d be the sum of the digits of m. The sum of the digits of m,m+1,...,m+19 are then k,k+1,k+2,...,k+9,k+1,k+2,...,k+10 respectively. One of the eleven consecutive integers, k,k+1,k+2,...,k+10 is divisible by eleven.
