Solution: By changing $(x, y, z)$ to $(-x, -y, -z)$, the condition (1) remains valid and the value of $x+y+z$ changes sign. Since the ordering of the numbers $x, y, z$ is irrelevant and they can not be of the same sign because of (1), we can without the loss of generality assume $x > 0, y > 0$, and $z < 0$, and find the maximum value of $V = |x + y + z|$.
We can transform (1) equivalently to
$$f(x) + f(y) = f(t), \quad \text{where} \quad t := -z > 0 \quad \text{and} \quad f(x) = x + \frac{1}{x},$$
with $x, y, t \in I := (1, \infty)$. In the following, we shall use the well known fact about the function $f$: It is an increasing bijection mapping $I$ onto $(2, \infty)$ (the trivial proof is omitted here). Applying this, from
$$f(t) = f(x) + f(y) = x + \frac{1}{x} + y + \frac{1}{y} > x + y + \frac{1}{x+y} = f(x+y)$$
we get $t > x + y$, hence $x + y + z = x + y - t < 0$. Therefore our aim is to maximize the positive expression $V = |x + y + z| = t - x - y$.
We will call valid every triple $(x, y, t) \in I^3$ satisfying $f(x) + f(y) = f(t)$. One of the valid triples is $(1, 1, t_0)$, with $t_0 > 1$ such that $f(t_0) = 4$, i. e., $t_0 = 2 + \sqrt{3}$. For this triple, we have $V = \sqrt{3}$. We shall show this is the desired maximum. The proof will be based on the following lemma: Whenever $(x, y, t)$ and $(x, y', t')$ are valid triples with the same first component $x$, and $y' < y$, we have $t - x - y < t' - x - y'$. If this lemma is true, then, with respect to the symmetry, the similar conclusion holds for the valid triples $(x, y, t)$ and $(x', y, t')$ with the same second component $y$. Hence, any valid triple $(x, y, t)$ can be replaced by $(x, 1, t')$ and then by $(1, 1, t_0)$, and the value of $V$ increases or remains the same during this process, concluding $\max V = \sqrt{3}$. (Moreover, this also implies that $(1, 1, 2+\sqrt{3})$ is the only valid triple for which the maximum value $V = \sqrt{3}$ is reached.)
To prove the lemma, notice that from
$$f(t) = f(x) + f(y), \quad f(t') = f(x) + f(y'),$$
and from $f(y') < f(y)$ (which is implied by the given assumption $y' < y$), we have $f(x) < f(t') < f(t)$, hence $1 < t' < t$. Therefore $1 < t'y' < ty$, and from
$$f(x) = f(t) - f(y) = (t - y) \left( 1 - \frac{1}{ty} \right) = f(t') - f(y') = (t' - y') \left( 1 - \frac{1}{t'y'} \right),$$
using the estimates
$$0 < 1 - \frac{1}{t'y'} < 1 - \frac{1}{ty},$$
we obtain $t - y < t' - y'$, which is equivalent to the desired $t - x - y < t' - x - y'$. Answer. The maximum value of $x + y + z$ is $\sqrt{3}$.

Second solution. We start in the same way as in the previous solution, reducing to the case $x \ge 1$, $y \ge 1$, and $z \le -1$ with maximizing $V = |x + y + z|$. From (1) we can express $z$ as the solution of the quadratic equation
$$z^2 + \left(x + \frac{1}{x} + y + \frac{1}{y}\right) z + 1 = 0.$$
Since the product of the roots of this equation equals 1, the value of $z$ lying outside of $(-1, 1)$ is, with respect to $x \ge 1$, $y \ge 1$, given by
$$z = \frac{1}{2} \left( - \left( x + \frac{1}{x} + y + \frac{1}{y} \right) - \sqrt{ \left( x + \frac{1}{x} + y + \frac{1}{y} \right)^2 - 4 } \right).$$
Therefore we are left to maximize
$$2V = 2|-x-y-z| = \left| \sqrt{ \left( x + \frac{1}{x} + y + \frac{1}{y} \right)^2 - 4 } - \left( x - \frac{1}{x} + y - \frac{1}{y} \right) \right|.$$
One can easily check that the expression inside of the last bars is positive. It remains
to show that
$$\sqrt{\left(x + \frac{1}{x} + y + \frac{1}{y}\right)^2 - 4} - \left(x - \frac{1}{x} + y - \frac{1}{y}\right) \le 2\sqrt{3}.$$
This inequality is equivalent (after rearranging, checking the positivity of both sides, canceling the square root, and simple calculations) to
$$\sqrt{\left(x + \frac{1}{x} + y + \frac{1}{y}\right)^2 - 4} \le 2\sqrt{3} + \left(x - \frac{1}{x} + y - \frac{1}{y}\right), \\ x^2 + y^2 + \sqrt{3}x + \sqrt{3}y \le 2xy + \sqrt{3}x^2y + \sqrt{3}xy^2,$$
which is true, since it is the sum of the inequalities
$$x^2 \le x^2 y, \quad y^2 \le xy^2, \quad \sqrt{3}x \le (\sqrt{3}-1)x^2y + xy, \quad \text{and} \quad \sqrt{3}y \le (\sqrt{3}-1)xy^2 + xy,$$
and these are trivially valid for $x, y \ge 1$.