4. Let's call a natural number "remarkable" if all its digits are different, it does not start with the digit 2, and by erasing some of its digits, the number 2018 can be obtained. How many different seven-digit "remarkable" numbers exist?
Answer: 1800.

Solution: To correctly count the number of options, it is necessary to follow the rule: before the digit 2, there must be one of the six digits $-3,4,5,6,7$ or 9. Let's assume for definiteness that this is 3, then two different digits from the remaining five ( $\frac{5 \cdot 4}{2}=10$ options) can be in the positions marked by dots: ..32..0..1..8.., i.e., either both numbers are in one interval $5 \cdot 2=10$ ways, or in different intervals, then $\frac{5 \cdot 4}{2} \cdot 2=20$ ways. Therefore, the total number of different options is $10 \cdot 30=300$. Similarly, the options for pairs $42,52,62,72$ and 92 are calculated, i.e., finally we get: $6 \cdot 300=$ 1800 options.

Comment: For a computational error with correct reasoning, deduct 1 point. For potentially useful but unimplemented ideas - 1-2 points.