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 .
Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$.
Obviously, $f(n)$ is monotonically decreasing. From the maximum value of $f(n)$, $f(1)<a-2007 \frac{1}{3}$, we get $a=2009$.