$\because f:\{1,2,3,4\} \rightarrow\{a \in \mathbb{Z}:|9| \leq 8\}$

and $f(n)+\frac{1}{n} f(n+1)=1$

$\Rightarrow n f(n)+f(n+1)=n \quad \ldots$ (i)

$\therefore f(1)+f(2)=1 \Rightarrow f(2)=1-f(1)$

But $f(1) \in[-8,8]$

Hence, $f(2) \in[-8,8] \Rightarrow f(1) \in[-7,8] \quad\ldots(\mathrm{A})$

and $2 f(2)+f(3)=2 \Rightarrow f(3)=2 f(1)$

$\therefore 2 f(1) \in[-8,8] \Rightarrow f(1) \in[-4,4] \quad\ldots(\mathrm{B})$

and $3 f(3)+f(4)=3 \Rightarrow f(4)=3-6 f(1)$

$\therefore f(1) \in\left[-\frac{5}{6}, \frac{11}{6}\right]\quad...(C)$

From (A), (B) and (C) : $f(1)=0$ or 1

$\therefore$ Only two functions are possible.