$\frac{2abc}{(a+b-c)(b+c-a)(c+a-b)}$ a positive integer

I am afraid your conjecture is not correct. The situation is quite complex. I have been studying such cubic representation problems for some time.

The basic identity \begin{equation*} \frac{2abc}{(a+b-c)(b+c-a)(c+a-b)}=N \end{equation*} is equivalent to finding points of infinite order on the elliptic curve \begin{equation*} v^2=u^3+2(2N^2-2N-1)u^2+(4N+1)u \end{equation*}

For specifically positive solutions the situation is even more complicated, in that we also need a point where the u-coordinate is negative.

Such situations are quite rare, but other examples come from $N=74, 218$ which give the results in the comment. A novel solution is for $N=250$ with $(97, 10051, 10125)$ as a solution.

We can also use multiples of the points in the elliptic-curve formulation to get larger solutions. For example, $N=26$ has the solution

$a=4739\,2819\,4344\,87$

$b=2634\,1867\,7932\,41$

$c=7228\,3008\,5646\,67$

and we can keep going getting larger and larger solutions.

There are more solutions, e.g. $(27, 1805, 1813)$ , $(115, 5239, 5341)$.