4. Find the maximum value of the natural number $a$ such that the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3 n+1}>2 a-5$ holds for all natural numbers $n$.

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4. Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3 n+1}\left(n \in \mathbf{N}^{+}\right)$
$$
f(n+1)-f(n)=\frac{1}{3 n+2}+\frac{1}{3 n+4}-\frac{1}{3 n+3}=\frac{2}{3(n+1)(3 n+2)(3 n+4)}>0 \text {, }
$$

Therefore, $f(n+1)>f(n)$, so $f(n)$ is an increasing function on $\mathbf{N}^{+}$, and $f(n)_{\min }=f(1)=\frac{13}{12}$. Thus, the necessary and sufficient condition for $f(n)>2 a-5$ to hold is $\frac{13}{12}>2 a-5 \Rightarrow a<\frac{73}{24}$, so $a_{\max }=3$.