Example 1 Write three integers on a blackboard, then erase one of them and replace it with the difference of 1 and the sum of the other two numbers. After repeating this process several times, the result is $(17,1999,2105)$. Could the three numbers initially written on the blackboard have been $(2,2,2)$ or $(3,3,3)$?
Due to the parity change relationship of the three numbers in the process described in the problem:
(even, even, even $) \rightarrow$ (even, even, odd $) \rightarrow$ (even, even, odd) (invariant), so $(2,2,2)$ cannot reach $(17,1999,2015)$ after several operations.
Since (odd, odd, odd) $\rightarrow$ (odd, odd, odd), $(3,3,3)$ could be the initial state of the operation process.
First, consider $(3,3,3)$ changing to $(17, a, b)(17<a<b)$ after several changes, thus $b=17+a-$ $1=a+16$. Clearly, $2015=1999+16$, fixing 17, choose the smallest $a$ such that $(17, a, a+16)$ can reach $(17,1999,2015)$ after several operations.

For $(17, a, a+16$ ), erasing $a$, after the operation, we get $(17, a+16, a+2 \times 16)$ (regardless of order), then erasing $a+16$, after the operation, we get $(17, a+2 \times 16, a+3 \times 16), \cdots$, after $n$ operations, we get $(17,17+16 n, a+$ $16(n+1))$. Since $1999=31+123 \times 16$, if we take $a=31$, then $(17,31,47)$, after 123 operations, can reach $(17,1999,2015)$.

Also, there is the following reverse inference: $(17,31,47) \leftarrow(15,17,31) \leftarrow(3,15,17) \leftarrow(3,13,15) \leftarrow(3,11$, 13) $\leftarrow(3,9,11) \leftarrow(3,7,9) \leftarrow(3,5,7) \leftarrow(3,3,5) \leftarrow(3,3,3)$, so $(17,31,47)$ can be reached from $(3,3,3)$ after 9 transformations, thus $(17,1999,2015)$ can be reached from $(3,3,3)$ after $123+9=132$ operations.