8. Given an even function $f: \mathbf{Z} \rightarrow \mathbf{Z}$ that satisfies $f(1)=1, f(2008) \neq 1$, and for any integers $a, b$, $f(a+b) \leqslant \max \{f(a), f(b)\}$, then the possible value of $f(2009)$ is
8. 1 Detailed Explanation: Since $f(2) \leqslant \max \{f(1), f(1)\}=1$, assume for a positive integer $k$ greater than or equal to 2, $f(k) \leqslant 1$, then $f(k+1) \leqslant \max \{f(k), f(1)\}=1$. Therefore, for all positive integers $n$, $f(n) \leqslant 1$. Since $f(x)$ is an even function, it follows that for all non-zero integers $n$, $f(n) \leqslant 1$. Because $f(2008) \neq 1$, we have $f(2008)<1$. Hence, $1=f(1)=f(2009+(-2008)) \leqslant$ $\max \{f(2009), f(-2008)\} \leqslant 1$, and since $f(-2008)=f(2008)<1$, it must be that $f(2009)=1$.