3. (10 points) Add the tens, hundreds, and thousands digits of 2015, and write the sum after the units digit of 2015, resulting in the natural number 20153; add the tens, hundreds, and thousands digits of the new number, and write the sum after the units digit of 20153, resulting in 201536; perform this operation 2 more times, resulting in 201536914. Continue this process for a total of 2015 times, resulting in a very large natural number. The sum of all the digits of this natural number is $\qquad$
【Analysis】According to the conditions given in the problem, after 16 operations, as shown in the figure above, the pattern of the numbers is found to be: starting from the 7th operation, the numbers are 11, 3, 3, 5, 7, and from the 12th operation, they are 11, 3, 3, 5, 7, these 5 numbers repeat.

Based on the entire pattern, we can deduce the number obtained after 2015 operations, and then find the sum.
【Solution】Solution: According to the conditions given in the problem, after 16 operations, as follows:

The pattern of the numbers is: starting from the 7th operation, the numbers are 11, 3, 3, 5, 7, and from the 12th operation, they are 11, 3, 3, 5, 7, these 5 numbers repeat,
then after 2015 operations: \((2015-6) \div 5=401 \cdots 4\), the number corresponding to the 2015th operation is 5; the sum of all natural numbers is:

The first 4 digits: \(2+0+1+5=8\),
The next 6 digits: \(3+6+9+1+4+1+6+6=36\),
The sum of the repeating numbers: \(1+1+1+3+3+5+7=21\), after repeating 401 times, the sum is \(401 \times 21=8421\), the remainder 4, the sum of the corresponding numbers is: \(1+1+1+3+3+5=14\),
Adding all the above numbers gives the sum of all natural numbers \(=8+36+8421+14=8479\).
Therefore: The answer should be: 8479.