, 2015

The number may be rewritten as
$$\frac{a}{f_1 \cdots f_n} (f_2 \cdots f_n + f_3 \cdots f_n + \cdots + f_{n-2}f_{n-1}f_n + f_{n-1}f_n + 1).$$
If this is an integer, then
$$f_n|a(f_2 \cdots f_n + f_3 \cdots f_n + \cdots + f_{n-2}f_{n-1}f_n + f_{n-1}f_n + 1).$$
Thus $f_n|a$. First consider the case where $n \ge 2$. Then
$$f_n \le a \le f_{n+1} < 2f_n.$$
Hence $a = f_n$. Then
$$\frac{1}{f_1 \cdots f_{n-1}} (f_2 \cdots f_n + f_3 \cdots f_n + \cdots + f_{n-2}f_{n-1}f_n + f_{n-1}f_n + 1)$$
is an integer. Thus $f_{n-1}|1$. Hence $n-1=0$ or $1$. This shows that the only possible values of $n$ are 1 and 2.
For $n=1$, $1=f_1 \le 1 \le f_2$ and $a/f_1 = 1/1 = 1$ is an integer.
For $n=2$, $f_2 \le 2 \le f_3$ and
$$2\left(\frac{1}{f_1} + \frac{1}{f_1 f_2}\right) = 2\left(1 + \frac{1}{2}\right)$$
is an integer.