From the fact that $\frac{bc(b+c)+ca(c+a)+ab(a+b)}{abc}$ is an integer, we have that $a|bc(b+c)$ and its cyclic variants. WLOG, $a \le b \le c$. We have either $a+b=c$ or $a+b = 2c$. If $a+b=2c$, $a=b=c=1$ so the value would be $6$. If $a+b=c$, the above value will be $3+2(\frac{a}{b}+\frac{b}{a})$. Since $\frac{a}{b}+\frac{b}{a}=\frac{n}{2}$, $\frac{a}{b}$ will be a solution to $x^2-\frac{n}{2}x+1$. Since the determinant of the quadratic is $\frac{n^2-16}{4}$, which needs to be a square of a rational. We get that $n=4,5$ and we get that the value is $7,8$. So in total, the possible values are $6,7,8$. This problem was in the Korean JMO Round 2 as well.

zschess wrote: Let $ a,b,c $ be pairwise coprime positive integers. Find all positive integer values of $$ \frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} $$ From the statement we have that $abc | ab(a + b) + bc(b + c) + ac(a + c)$ so we get that: $c | a + b$,$a | b + c$,$b | c + a$ so $a,b,c|a + b + c$ $\implies$ $abc | a + b + c$ so $a + b + c \ge abc$. By simetry $a \ge b \ge c$. But $c < 3$ so we have $2$ cases: $1)$ If $c = 1$ the $ab | a + b + 1$,obviously $b < 3$. If $b = 1$ then $a \in \{1,2\}$. If $b = 2$ then $a \in \{3\}$. $2)$ If $c = 2$ then $ab | a + b + 2$,again obviously $b < 3$. If $b = 2$ then $2a | a + 4$,contadiction $a \in \{ 2,4 \}$ and $(a,b) = 2$. So $(a,b,c) \in \{(1,1,1),(2,1,1),(3,2,1)\}$.