1. $16 N$ is a 1990-digit number, and it is a multiple of 9. The sum of the digits of $N$ is $N_{1}, N_{1}$'s sum of the digits is $N_{2}, N_{2}$'s sum of the digits is $N_{3}$. Find $N_{3}$.
[Solution] If a number is a multiple of 9, then the sum of its digits is also a multiple of 9. Since $N$ is a multiple of 9, $N_{1}$ is a multiple of 9 not greater than $9 \times 1990$; since $9 \times 1990$ is a 5-digit number, $N_{2}$ is a multiple of 9 not greater than $9 \times 5=45$, thus, we get $N_{3}=9$.