Let the side lengths of the triangle be integers $a$, $b$, $c$. Without loss of generality we may assume that $c \ge a$ and $c \ge b$. Suppose that the perimeter of the triangle $a+b+c$ is divisible by double of the longest side length $2c$. Since $0 < a+b+c \le 3c < 2 \cdot 2c$, the perimeter $a+b+c$ can be divisible by $2c$ only in the case when $a+b+c = 2c$. But then $a+b = c$, which violates the triangle inequality $a+b > c$.