4. (8 points) There is a sequence, the first term is 12, the second term is 19, starting from the third term, if the sum of its previous two terms is odd, then this term equals the sum of the previous two terms, if the sum of the previous two terms is even, this term equals the difference of the previous two terms (the larger number minus the smaller number). Then, in this sequence, the $\qquad$ term first exceeds 2016.
【Solution】Solution: According to the problem, we have:
The sequence is $12,19,31,12,43,55,12,67,89,12,101,113,12 \cdots$
By summarizing the pattern, every three numbers form a group. If we remove all the 12s, we find that it is an arithmetic sequence with the first term 19 and a common difference of 12.
$$
(2016-19) \div 12=166 \cdots 5
$$

This indicates that $19+167 \times 12=2023$.
This means it is the 168th term of the arithmetic sequence.
Since one number is undercounted in each group, the number of terms is $168 \div 2 \times 3=252$.
Therefore, the answer is: 252