Expanding the above expression $a(\frac{f_2f_3..f_n+f_3..f_n+\cdots f_{n-1}f_n+f_n+1}{f_1f_2..f_n}$ $\implies f_n|a(f_2f_3..f_n+f_3..f_n+\cdots +f_{n-1}f_n+f_n+1)$ $\implies f_n|a$ for $n\geq 3$, and moreover, $2f_n>f_{n+1}$ as $f_{n+1}=f_n+f_{n-1}$ Hence we can deduce $a=f_n$ for $n\geq 3$ We get that a gets cancelled. And $f_{n-1}|f_n+1$

Now using our claim we get n is even. We also get$f_{n-2}|f_{n-1}f_n+f_n+1$ Using again our claim $f_{n-1}^2=f_nf_{n-2}+(-1)^{n-1}$ as n is even we get $f_{n-1}^2\equiv -1(mod f_{n-2})$$-(1)$

But we know that $\boxed{(f_n,f_{n+1})=1}$ So we are only left with cases $n\in1,2,3$ When $n=1$ we have $\boxed{a\in1,2}$. When $n=2$ we have $\boxed{a\in2}$ When $n=3$ we have $\boxed{a\in3}$$\blacksquare$

$$a \left( \frac{1}{f_1}+\frac{1}{f_1f_2}+\cdots+\frac{1}{f_1f_2...f_n} \right) = a \left( \frac{f_2 f_3 \ldots f_n+ f_3f_4 \ldots f_n + \ldots + f_n + 1}{f_1f_2...f_n} \right)$$ so $$f_1f_2...f_n \mid a(f_2 f_3 \ldots f_n+ f_3f_4 \ldots f_n + \ldots + f_n + 1).$$ In particular $$f_n \mid a(f_2 f_3 \ldots f_n+ f_3f_4 \ldots f_n + \ldots + f_n + 1)$$ but since $f_2 f_3 \ldots f_n+ f_3f_4 \ldots f_n + \ldots + f_n + 1 \equiv 1 \pmod {f_n}$ implies $\gcd(f_2 f_3 \ldots f_n+ f_3f_4 \ldots f_n + \ldots + f_n + 1,\ f_n) = 1$, we must have $f_n \mid a$. But $a \leq f_{n+1} < 2 f_n$, so we must have $a = f_n$. This implies $$f_1f_2...f_{n-1} \mid f_2 f_3 \ldots f_n+ f_3f_4 \ldots f_n + \ldots + f_n + 1.$$ so $f_{n-1} \mid 1$ so $n-1 = 0 \text{ or } 1$. We check that $n = 1,2$ both yield solutions.