Answer 6.
Analysis From the problem, we have
$$
f(x)=x+\frac{4}{x}-1 \in\left[3,15 \frac{1}{4}\right]
$$

Therefore, when $n$ is as large as possible, the scenario is $3+3+3+3+3 \frac{1}{4}=15 \frac{1}{4}$, at this time $n=6$.
11. Given the function $f(x)=x+\frac{4}{x}-1$, if there exist $x_{1}, x_{2}, \cdots, x_{n} \in\left[\frac{1}{4}, 4\right]$, such that $f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n-1}\right)=f\left(x_{n}\right)$, then the maximum value of the positive integer $n$ is $\qquad$.