16. (15 points) Observe the following operations:
If $\overline{\mathrm{abc}}$ is a three-digit number, since $\overline{\mathrm{abc}}=100 a+10 b+c=99 a+9 b+(a+b+c)$,
Therefore, if $a+b+c$ is divisible by 9, $\overline{\mathrm{abc}}$ is divisible by 9.
This conclusion can be extended to any number of digits.
Using the above conclusion, solve the following problems:
(1) $N$ is a 2011-digit number, with each digit being 2, find the remainder when $N$ is divided by 9.
(2) $N$ is an $n$-digit number, with each digit being 7, and $n$ is a number that leaves a remainder of 3 when divided by 9. Find the remainder when $N$ is divided by 9.
【Solution】Solution: (1) $2011 \times 2=4022$;
$$
4022 \div 9=446 \cdots 8 \text {, }
$$

Therefore, when $N$ is divided by 9, the remainder is 8;
(2) The sum of the digits of the natural number $N$ is $7n$; since $n \div 9$ has a remainder of 3, we can assume $n=9k+3$, then $7n=7(9k+3)=63k+21=(63k+18)+3=9(7k+2)+3$;

Thus, the sum of the digits of $\mathbf{N}-3$ is $7n-3=9(7k+2)$, which is divisible by 9, so $\mathbf{N}-3$ is divisible by 9, hence the remainder when $N$ is divided by 9 is also 3.

Answer: (1) The remainder when $N$ is divided by 9 is 8, (2) The remainder when $N$ is divided by 9 is 3.