4. There is a four-digit integer. By adding a decimal point in front of one of its digits, and then adding this number to the four-digit integer, the sum is 2000.81. Find this four-digit number.
1981
4.【Solution】Since the result has two decimal places, the decimal point cannot be added before the units digit. If the decimal point is added before the tens digit, the resulting number is one percent of the original four-digit number. Adding this to the original four-digit number, the result 2000.81 should be 1.01 times the original four-digit number, so the original four-digit number is $2000.81 \div 1.01=1981$.

Similarly, if the decimal point is added before the hundreds digit, the result 2000.81 should be 1.001 times the original four-digit number; if the decimal point is added before the thousands digit, the result 2000.81 should be 1.0001 times the original four-digit number. However, $(2000.81 \div 1.001)$ and $(2000.81 \div 1.0001)$ are not integers, so 1981 is the only possible answer.

Answer: The four-digit number is 1981.

【Solution】Notice that in the original four-digit number, the digits 8 and 1 must appear in sequence. The decimal point cannot be added before the units digit; it also cannot be added before the thousands digit, otherwise the original four-digit number would have to be 8100, which is greater than 2000.81.

Whether the decimal point is added before the tens digit or the hundreds digit, the resulting number is greater than 1 and less than 100. This number plus the original four-digit number equals 2000.81, so the original four-digit number must be less than 2000 but greater than 1900, indicating that its first two digits must be 1 and 9. Since it also has the digits 8 and 1 in sequence, it can only be 1981.