13. (25 points) As shown in Figure 1, with the right triangle $\triangle ABC (\angle C = 90^\circ)$, squares $CADE$, $BCFG$, and $ABHI$ are constructed outward on the sides $CA$, $CB$, and $AB$ respectively. Let the lengths of sides $CB$ and $CA$ be $a$ and $b$ respectively, and the area of the convex hexagon $DEFGHI$ be $S$. Does there exist positive integers $a$ and $b$ such that $S = 2016$? If so, find all such positive integers $a$ and $b$; if not, explain why.
13. As shown in Figure 5, extend $BA$ to point $B'$ such that $B'A = AB$, and connect $B'C$.
It is easy to prove that $\triangle B'AC \cong \triangle IAD$.
Thus, $S_{\triangle IAD} = S_{\triangle B'AC} = S_{\triangle ABC}$.
Similarly, $S_{\triangle ECF} = S_{\triangle ABC} = S_{\triangle HBG}$.
Therefore, $S = 4 \times \frac{1}{2} ab + 2(a^2 + b^2)$
$$
= 2(a^2 + b^2 + ab).
$$

Assume there exist positive integers $a, b$ such that $S = 2016$, i.e.,
$$
a^2 + b^2 + ab = 2^4 \times 3^2 \times 7.
$$

It is easy to see that $a, b$ are both even (otherwise, if $a, b$ are both odd or one is odd and the other is even, then $a^2 + ab + b^2$ would be odd, which is a contradiction).
Let $a = 2a_1, b = 2b_1$. Substituting into equation (1) gives
$$
a_1^2 + b_1^2 + a_1 b_1 = 2^2 \times 3^2 \times 7.
$$

Similarly, $a_1, b_1$ are both even.
Let $a_1 = 2x, b_1 = 2y$.
Thus, $x^2 + y^2 + xy = 3^2 \times 7$.
Assume without loss of generality that $x \leq y$, then $3^2 \times 7 \geq 3x^2$. Hence, $x^2 \leq 21$.
Since $x, y$ are positive integers, then $x = 1, 2, 3, 4$.
Substituting into equation (2), after inspection, only when $x = 3$, $y^2 + 3y = 54 \Rightarrow y = 6$ or -9 (discard).
Thus, $(x, y) = (3, 6), (6, 3)$
$$
\Rightarrow (a, b) = (12, 24), (24, 12).
$$
(Provided by Li Changyong)