3. For a natural number $n$, we say it is convenient if the digits that participate in its notation can be divided into two groups such that the sums of the digits in those groups are equal to each other (for example, the number 121 is convenient because we can put the two ones in one group and the digit 2 in the other, and then $1+1=2$; the number 2019 is not convenient because, with any partition, the group containing the digit 9 will have a larger sum than the other group).

a) Determine the smallest natural number $n$ such that both numbers $n$ and $n+1$ are convenient.

b) Do there exist three consecutive natural numbers that are all convenient?
3. a) Let $n$ be a suitable natural number such that the number $n+1$ is also suitable. It is clear that the sum of the digits of any suitable number must be even. Therefore, the last digit of the number $n$ must be 9 (because otherwise, the sum of the digits of the number $n+1$ would be 1 greater than the sum of the digits of the number $n$, and both of these sums could not be even). The only two-digit suitable number that ends with the digit 9 is the number 99, but we have $99+1=100$, and 100 is not a suitable number; hence, the desired number $n$ must be at least three digits. If we write $n=\overline{a b 9}$, then we have $n+1=\overline{a(b+1) 0}$ (obviously, $b<9$, because $\overline{a 99}$ cannot be suitable). For both $n$ and $n+1$ to be suitable, it must hold that $a+b=9$ and $a=b+1$, from which it follows that $a=5$ and $b=4$. Therefore, the smallest desired number $n$ is 549.

b) As we obtained in part a), for both $n$ and $n+1$ to be suitable numbers, $n$ must end with the digit 9, and similarly, for both $n+1$ and $n+2$ to be suitable numbers, $n+1$ must end with the digit 9. This is clearly not possible, so it follows that the triplet sought in the problem does not exist.