By the given condition, $b \mid f_1 - 2a$, that is, $b \mid 1 - 2a$. But since $b > a$, we get $b = 2a - 1$. Moreover, for every positive integer $n$, we have
$$b \mid f_n - 2n a^n, \quad b \mid f_{n+1} - 2(n+1)a^{n+1}, \quad b \mid f_{n+2} - 2(n+2)a^{n+2}. \quad (1)$$
From these three relations, together with $f_{n+2} = f_{n+1} + f_n$ and $b = 2a-1$, we know that
$$b \mid (n + 2)a^{n+2} - (n + 1)a^{n+1} - n a^n.$$
Also, since $b = 2a - 1$ is necessarily coprime to $a$, we obtain
$$b \mid (n + 2)a^2 - (n + 1)a - n. \quad (2)$$
Substituting $n+1$ for $n$ in (2) gives
$$b \mid (n + 3)a^2 - (n + 2)a - (n + 1). \quad (3)$$
Subtracting the right-hand sides of (2) and (3), we obtain
$$\begin{aligned} & b \mid a^2 - a - 1, \\ \text{that is,}\quad & (2a - 1) \mid a^2 - a - 1, \\ & (2a - 1) \mid 4a^2 - 4a - 4. \end{aligned}$$
Since $4a^2 - 4a - 4 = (2a - 1)^2 - 5$, it follows that $2a - 1$ divides $-5$, so $2a - 1 = 1$ or $5$, which yields only one solution $a = 3, b = 2a - 1 = 5$. (The other solution $a = 1, b = 2a - 1 = 1$ does not satisfy $b > a$, so it is discarded.)

Finally we check sufficiency, that is, that $(a, b) = (3, 5)$ satisfies the given condition; that is, for every positive integer $n$, $f_n - 2n \cdot 3^n$ is indeed divisible by $5$. When $n = 1, 2$, $f_1 - 2 \cdot 1 \cdot 3 = 1 - 6 = -5$, $f_2 - 2 \cdot 2 \cdot 3^2 = 1 - 36 = -35$ are indeed both divisible by $5$.

Now suppose the conclusion holds for $n = k, k + 1$, that is, $f_k - 2k \cdot 3^k$ and $f_{k+1} - 2(k + 1) \cdot 3^{k+1}$ are both divisible by $5$. Then $5$ also divides
$$(f_{k+1} - 2(k+1) \cdot 3^{k+1}) + (f_k - 2k \cdot 3^k) = f_{k+2} - 2 \cdot 3^k (4k + 3).$$

Thus, for $5$ to divide $f_{n+2} - 2(k+2) \cdot 3^{k+2}$, this condition is equivalent to
$$9(k + 2) \equiv 4k + 3 \pmod{5}.$$
But this is equivalent to $5$ dividing $5k+15$, which clearly holds. Hence, by mathematical induction, the problem is proved.