(4) The smallest positive integer $a$ that makes the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}<a-2007 \frac{1}{3}$ hold for all positive integers $n$ is $\qquad$.
(4) 2009 Hint: Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$. Clearly, $f(n)$ is monotonically decreasing. Then, by the maximum value of $f(n)$, $f(1)<a-2007 \frac{1}{3}$, and $a$ is a positive integer, we get $a=2009$.