10.5. Represent the number 1000 as the sum of the maximum possible number of natural numbers, the sums of the digits of which are pairwise distinct.
Answer: 19.

Solution: Note that the smallest natural number with the sum of digits A is a99..99, where the first digit is the remainder, and the number of nines in the record is the incomplete quotient of the division of A by 9. From this, it follows that if A is less than B, then the smallest number with the sum of digits A is also less than the smallest number with the sum of digits B. Consider the smallest natural numbers with the sums of digits $1,2,3, \ldots, 19,20$, they are respectively $a_{1}=1, a_{2}=2, \ldots a_{9}=9, a_{10}=19, a_{11}=29, \ldots a_{18}=99, a_{19}=199, a_{20}=299$. The sum of the first 19 of them is 775, and the remainder of its division by 9 is 1, which is the same as the remainder of the division of 1000 by 9. Increase this sum by 225, taking 234 instead of 9, with the same sum of digits, and we get a representation of 1000 as the sum of 19 numbers $1,2,3,4,5,6,7,8,19,29,39,49,59,69,79,89,99,199,234$ with different sums of digits from 1 to 19.

Suppose that 1000 can be represented as the sum of $n \geq 20$ natural numbers $b_{1}, b_{2}, \ldots b_{n}$ with different sums of digits $s\left(b_{1}\right)<s\left(b_{2}\right)<\ldots<s\left(b_{n}\right)$. In this case, $s\left(b_{k}\right) \geq k, k=1,2, \ldots n,$ and, consequently, $b_{k} \geq a_{k}, k=1,2, \ldots n$. Therefore, $1000=b_{1}+b_{2}+\ldots+b_{n} \geq a_{1}+a_{2}+\ldots+a_{n} \geq a_{1}+a_{2}+\ldots+a_{20}=107$ - a contradiction.

Grading criteria: Finding a representation of 1000 as the sum of 19 numbers with different sums of digits: 3 points. Proof of the maximality of 19: 4 points.