Example 6 For the function $y=(n-1) t^{2}-10 t+10, t \in(0,4]$, does there exist a positive integer $n$ such that $0<y \leqslant 30$ always holds?
Solve $0<(n-1) t^{2}-10 t+10 \leqslant 30 \Leftrightarrow 0<n t^{2}-t^{2}-10 t+10 \leqslant 30 \Leftrightarrow-\frac{10}{t^{2}}+$ $\frac{10}{t}+1<n \leqslant \frac{20}{t^{2}}+\frac{10}{t}+1, \frac{1}{t} \in\left[\frac{1}{4},+\infty\right) \Leftrightarrow-10\left(\frac{1}{t}-\frac{1}{2}\right)^{2}+\frac{7}{2}<n \leqslant$ $20\left(\frac{1}{t}+\frac{1}{4}\right)^{2}-\frac{1}{4}, \frac{1}{t} \in\left[\frac{1}{4},+\infty\right) \Leftrightarrow\left[-10\left(\frac{1}{t}-\frac{1}{2}\right)^{2}+\frac{7}{2}\right]_{\max }<n \leqslant$ $\left[20\left(\frac{1}{t}+\frac{1}{4}\right)^{2}-\frac{1}{4}\right]_{\text {min }}, \frac{1}{t} \in\left[\frac{1}{4},+\infty\right)$. Therefore, $3.5<n \leqslant \frac{19}{4}$, which means there exists $n=4$ such that $0<y \leqslant 30$ always holds.