1. Given $f(x)=\frac{10}{x+1}-\frac{\sqrt{x}}{3}$. Then the set
$$
M=\left\{n \in \mathbf{Z} \mid f\left(n^{2}-1\right) \geqslant 0\right\}
$$

has $\qquad$ elements.
,- 1.6 .
From the problem, we know that $f(x)$ is monotonically decreasing on the interval $[0,+\infty)$, and $f(9)=0$.
Then $f\left(n^{2}-1\right) \geqslant f(9) \Rightarrow 1 \leqslant n^{2} \leqslant 10$.
Thus, the number of elements in set $M$ is 6.