10. There is a six-digit number, the sum of its digits is divisible by 26. When this six-digit number is increased by 1, the sum of the digits of the resulting number is also divisible by 26. The smallest six-digit number that satisfies the above conditions is $\qquad$
10. 898999.

Obviously, this six-digit number will have a carry when 1 is added.
To make the sum of the digits a multiple of 26, at least three positions need to carry over. Therefore, the last three digits of this six-digit number are all 9, and the sum of the digits of the first three is 25.
Thus, the smallest number that meets the condition is 898999.