Example 1 Write two integers on the blackboard, then erase one of them and replace it with the sum or difference of the other two numbers. If this process is repeated several times, and the final result is (17.1999, 2015), then - what could the initial numbers on the blackboard be: (1)(2,2,2); (2)(3,3,3).
Solution: Due to the changes in the parity of the three numbers during the operation process as follows:
(even, even, even $) \rightarrow$ (even, even, odd $) \rightarrow$ (even, even, odd),

thus, (2,2,2) cannot reach $(17,1999,2015)$ after several operations.

Since (odd, odd, odd) $\rightarrow$ (odd, odd, odd) $\rightarrow$ (odd, odd, odd).

then $(3,3,3)$ could be the initial state of the operation process.
If we first consider $(3,3,3)$ transforming into $(17, a, b)(17,17$, such that (17, $a$, $a+16)$ can reach (17,1999, $2015$) after several operations.

For $(17, a, a+16)$, if we erase $a+16$, the operation will remain unchanged, so we should consider erasing $a$. After the operation, we get
$(17, a+16, a+2 \times 16)$.
Then, if we erase $a+16$, after the operation, we get $(17, a+2 \times$ $16, a+3 \times 16), \cdots \cdots$, after $n$ operations, we get $(17, a$ $+n \times 16, a+(n+1) \times 16)$.

Since $1999=31+123 \times 16$, if we take $a=31$, (17,31,47) needs 123 operations of erasing the middle number to reach (17,1999,2015).
Also, there is the following recursive relationship:
$(17,31,47) \leftarrow(15,17,31) \leftarrow(3,15,17) \leftarrow$ $(3,13,15) \leftarrow(3,11,13) \leftarrow(3,9,11) \leftarrow(3,7,9)$ $\leftarrow(3,5,7) \leftarrow(3,3,5) \leftarrow(3,3,3)$,

we know that $(17,31,47)$ can be reached from $(3,3,3)$ after 9 transformations. Therefore, $(17,1999,2015)$ can be reached from $(3,3,3)$ after 123 $+9=132$ transformations.