10. (12 points) The natural numbers $1, 2, 3, \cdots, 2014, 2015, 2016$ are arranged in a circle clockwise. Starting from the number 1, the following operations are performed clockwise:
First step: Cross out 1, keep 2;
Second step: Cross out $3, 4$, keep 5;
Third step: Cross out $6, 7, 8$, keep 9;
Fourth step: Cross out $10, 11, 12, 13$, keep $14; \cdots$;
That is, in the nth step, first cross out n numbers, then keep 1 number, and continue this operation until all numbers are crossed out.
What is the last number to be crossed out? $\qquad$
【Solution】Solution: According to the problem, we know:
The numbers retained in the first round are $2,5,9, \cdots$ Therefore, the maximum number retained in the first round is:
$$
\begin{array}{l}
2+3+4+\cdots+n \\
=\frac{2+n}{2} \times(n-1)
\end{array}
$$

When $n=63$, the sum of the sequence is 2015. This indicates that 2015 is a retained number.
At this point, not all numbers have been crossed out and further crossing is required. However, since it is a circle, the order of crossing continues as 2016, $2,5,9 \cdots$, this being the 63rd operation, 2015 is the last one to be crossed out.

Therefore, the answer is: 2015.