12. (5 points) From the 200 natural numbers from 1 to 200, how many numbers must be selected to ensure that there are at least 2 numbers whose sum is a multiple of 5?
【Analysis】First, classify the numbers according to their remainders when divided by 5, which are $0,1,2,3,4$. Then, numbers with remainders of 1 and 4 cannot appear together, and numbers with remainders of 2 and 3 cannot appear together. Numbers with a remainder of 0 can only appear once.

【Solution】Solution: According to the problem:
First, classify the numbers based on their remainders when divided by 5. We only need to consider numbers with remainders of 1 and 2.
$200 \div 5=40$ (groups), each group has two numbers with remainders of 1 and 2.
$40 \times 2=80$ (numbers); then find one number with a remainder of 0. Adding one more number will ensure that there are 2 numbers whose sum is a multiple of 5.
$$
80+1+1=82 \text { (numbers) }
$$

In summary: At least 82 numbers must be selected to ensure that there are 2 numbers whose sum is a multiple of 5.