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 $\qquad$.
Let $a_{n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}, a_{n+1}=\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2 n+3}$, then $a_{n+1}-a_{n}=\frac{1}{2 n+2}+\frac{1}{2 n+3}-\frac{1}{n+1}a_{1}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6} \Rightarrow a>2007 \frac{1}{3}+\frac{5}{6}=2008 \frac{1}{6} \Rightarrow a \geqslant 2009$. Therefore, the smallest positive integer $a$ is 2009.