Integers...

Notice that

$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{y}{x}+\frac{x}{z}+\frac{z}{y}=(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3$

so this expression must be an integer if $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is an integer. Assume that $x,y$ and $z$ are $k-1, k,$ and $k+1$ in some order. Then combining denominators and simplifying, we get that

$(k-1+k+k+1)\left(\frac{1}{k-1}+\frac{1}{k}+\frac{1}{k+1}\right)$

$=3k\left(\frac{3k^2-1}{k^3-k}\right)$

$=\frac{9k^2-3}{k^2-1}$

$=9+\frac{6}{k^2-1}$

must be an integer or $\frac{6}{k^2-1}$ must be an integer. Checking the positive divisors of $6$ and noting $k$ must be positive, we get that $k^2-1=3$ or $k=2$ . $x+y+z=3k=\boxed{6}$ .

I must admit, I actually guessed at the correct answer pretty easily, but spent some time today looking at this and got a proof together. It's good on paper, but need to figure out how to type it up. Haven't submitted solutions on here before. It's clever, though!