13. B. As shown in Figure 2, in $\triangle ABC$, the lengths of the three sides are $BC = a$, $CA = b$, $AB = c$, where $a$, $b$, and $c$ are all integers, and the greatest common divisor of $a$ and $b$ is 2. $G$ and $I$ are the centroid and incenter of $\triangle ABC$, respectively, and $\angle GIC = 90^{\circ}$. Find the perimeter of $\triangle ABC$.
13. B. As shown in Figure 8, extend $G I$, intersecting sides $B C$ and $C A$ at points $P$ and $Q$ respectively. Let the projections of the centroid $G$ on sides $B C$ and $C A$ be $E$ and $F$, and the inradius of $\triangle A B C$ be $r$. The lengths of the altitudes from $B C$ and $C A$ are $h_{a}$ and $h_{b}$ respectively. It is easy to see that $C P = C Q$.
From $S_{\triangle P Q C} = S_{\triangle G P C} + S_{\triangle C Q C}$, we get
$$
2 r = G E + G F = \frac{1}{3}\left(h_{a} + h_{b}\right),
$$

which means $2 \cdot \frac{2 S_{\triangle A B C}}{a + b + c} = \frac{1}{3}\left(\frac{2 S_{\triangle B B C}}{a} + \frac{2 S_{\triangle A B C}}{b}\right)$.
Thus, $a + b + c = \frac{6 a b}{a + b}$.
Since the centroid $G$ and the incenter $I$ of $\triangle A B C$ do not coincide, $\triangle A B C$ is not an equilateral triangle, and $b \neq a$. Otherwise, if $a = b = 2$, then $c = 2$, which is a contradiction.
Assume without loss of generality that $a > b$.
Since $(a, b) = 2$, let $a = 2 a_{1}$ and $b = 2 b_{1}$, with $\left(a_{1}, b_{1}\right) = 1$. Therefore, $\frac{6 a b}{a + b} = \frac{12 a_{1} b_{1}}{a_{1} + b_{1}}$ is an integer. Hence, $\left(a_{1} + b_{1}\right) \mid 12$, which means $(a + b) \mid 24$.

Therefore, only when $a = 14$ and $b = 10$, we get $c = 11$, satisfying the conditions.
Thus, the perimeter of $\triangle A B C$ is 35.