6. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, \sqrt{\frac{1}{a_{n}^{2}}+4}=\frac{1}{a_{n+1}}$, and let $S_{n}=\sum_{i=1}^{n} a_{i}^{2}$. If $S_{2 n+1}-S_{n} \leqslant \frac{t}{30}$ holds for any $n$ $\left(n \in \mathbf{N}_{+}\right)$, then the minimum value of the positive integer $t$ is ( ).
(A) 10
(B) 9
(C) 8
(D) 7
6. A.

From the given $\frac{1}{a_{n+1}^{2}}-\frac{1}{a_{n}^{2}}=4$, we get $a_{n}^{2}=\frac{1}{4 n-3}$. Let $g(n)=S_{2 n+1}-S_{n}$. Then $g(n)-g(n+1)=a_{n+1}^{2}-a_{2 n+2}^{2}-a_{2 n+3}^{2}$
$$
=\frac{1}{4 n+1}-\frac{1}{8 n+5}-\frac{1}{8 n+9}>0,
$$

which means $g(n)$ is a decreasing function.
Therefore, $S_{2 n+1}-S_{n} \leqslant g(1)=\frac{14}{45} \leqslant \frac{t}{30}$.
Thus, $t \geqslant \frac{28}{3}$. Hence, the minimum value of $t$ is 10.