a) Given six natural numbers. All of them are distinct and their sum is 22. Find these numbers and prove that there are no others.

b) The same question for 100 numbers, which sum up to 5051.
Let's arrange the numbers in ascending order. Then it is obvious that each number will be greater than its position number. Let's find the sum of the position numbers of all numbers: a) $1+2+3+4+5+6=21$; b) $1+2+\ldots+100=5050$. (The last sum can be calculated as follows: $(1+100)+(2+99)+\ldots+(50+51)=50 \cdot 101=5050$.) In both cases, this sum is one less than the sum of the numbers themselves. Therefore, one number is one greater than its position number, and the rest are equal to it. The number that is greater than its position number can only be the last one. Indeed, if some number is greater than its position number, then all subsequent numbers are also greater than their position numbers. Therefore, the required numbers in part a) are $1,2,3,4,5,7$; and in part b) $-1,2, \ldots, 99,101$.