6. The smallest natural number $a$ that makes the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$ $<a-1995 \frac{1}{3}$ hold for all natural numbers $n$ is $\qquad$.
6. 1997.

Let $\dot{f}(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$, then $f(n+1)-f(n)=\frac{-1}{(2 n+3)(2 n+2)}<0$, which means $f(n)$ is a decreasing function when $n \geq 0$. Therefore, $f(1)$ is the maximum value. To ensure the inequality holds for all $n \in N$, the maximum value of $f(n)$ must be less than: $a-1995 \frac{1}{3}$, i.e.,
$$
a-1995 \frac{1}{3}>f(1)=\frac{5}{6} \text {. }
$$

Thus, $a>1996 \frac{1}{6}$, and the smallest natural number $a$ is 1997.