5. In the arithmetic sequence $\left\{a_{n}\right\}$, $a_{2}=5, a_{6}=21$, let the sum of the first $n$ terms of the sequence $\left\{\frac{1}{a_{n}}\right\}$ be $S_{n}$. If $S_{2 n+1}-S_{n} \leqslant \frac{m}{15}$ holds for any $n \in \mathbf{N}_{+}$, then the minimum value of the positive integer $m$ is $\qquad$.
$$
d=\frac{a_{6}-a_{2}}{6-2}=4 \Rightarrow a_{n}=4 n-3,
$$

Let $A_{n}=S_{2 n+1}-S_{n}=\frac{1}{4 n+1}+\frac{1}{4 n+5}+\cdots+\frac{1}{8 n+1}, A_{1}=\frac{1}{5}+\frac{1}{9}$,
$A_{n+1}-A_{n}=\frac{1}{8 n+5}+\frac{1}{8 n+9}-\frac{1}{4 n+1}<0 \Rightarrow A_{n+1}<A_{n} \Rightarrow$ the sequence $\left\{A_{n}\right\}$ is a decreasing sequence, its maximum term is $A_{1}=\frac{1}{5}+\frac{1}{9}=\frac{14}{45}$. Therefore, $\frac{m}{15} \geqslant \frac{14}{45} \Rightarrow 3 m \geqslant 14 \Rightarrow m \geqslant 5$.
Thus, the smallest positive integer $m$ is 5.