13. (15 points) A sequence of numbers is first written on the blackboard: $1,2,3, \cdots, 100$. If each time the first 6 numbers are erased, and the sum of these 6 erased numbers is written at the end of the sequence, a new sequence of numbers is obtained. The same operation is repeated until fewer than 6 numbers remain on the blackboard. Ask:
(1) What is the sum of the numbers left on the blackboard at the end?
(2) What is the last number written?
【Analysis】First, analyze the first question: Erase $1,2,3,4,5,6$ but write down 21, the sum of the numbers remains unchanged. The sum of the remaining numbers is the sum of all the numbers. In the second question, it is found that the numbers form an arithmetic sequence, which can be enumerated.

【Solution】Solution: According to the problem:
(1) Erase $1,2,3,4,5,6$ but write down 21, the sum of the numbers remains unchanged.

The final sum of the numbers is $1+2+3+\cdots+100$, which is 5050.
(2) The numbers erased the first time are $1,2,3,4,5,6$ and the number written is 21. The numbers erased the second time are 7, 8, 9, 10, 11, 12, and the number written is 57. The difference between 21 and 57 is 36.
$100 \div 6=16 \cdots 4$. This means 96 numbers are erased and 16 numbers are added, forming an arithmetic sequence with the first term 21 and a common difference of 36. There are 20 numbers in total.

These 20 numbers are: $97,98,99,100,21,57,93,129,165,201,237,273,309,345,381,417,453,489,525,561$.

Then $20 \div 6=3 \cdots 2$. This means the last two numbers remain, and 3 new numbers are added. The last numbers written are 309, 345, 381, 417, 453, 489, with a sum of 2394.

Answer: (1) The sum of the remaining numbers on the blackboard is 5050. (2) The last number written is 2394.