1. The smallest natural number $a$ that makes the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}<u-2000 \frac{1}{2}$ hold for all $n \in \mathbf{N}$ is $\qquad$
1. 2002

Consider $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$ is monotonically decreasing, so $a>2000 \frac{1}{2}+f(1)=2001 \frac{1}{3}$. Therefore, $a_{\min }=2002$.