Real numbers $x,y,z$ satisfy $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+x+y+z=0$$ and none of them lies in the open interval $(-1,1)$. Find the maximum value of $x+y+z$.

[i]Proposed by Jaromír Šimša[/i]
Given the equation:
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + x + y + z = 0
\]
and the condition that none of \(x, y, z\) lies in the open interval \((-1, 1)\), we need to find the maximum value of \(x + y + z\).

1. **Analyze the function \(f(t) = t + \frac{1}{t}\)**:
   - The second derivative of \(f(t)\) is \(f''(t) = \frac{2}{t^3}\).
   - Since \(f''(t) > 0\) for \(t > 0\) and \(f''(t) < 0\) for \(t < 0\), \(f(t)\) is convex for \(t > 0\) and concave for \(t < 0\).

2. **Consider the signs of \(x, y, z\)**:
   - Since \(x, y, z\) cannot all have the same sign, assume \(x\) is negative and \(y, z\) are positive.
   - Let \(x = -a\) where \(a > 1\). Then the equation becomes:
     \[
     -\frac{1}{a} + \frac{1}{y} + \frac{1}{z} - a + y + z = 0
     \]
     Simplifying, we get:
     \[
     \frac{1}{y} + \frac{1}{z} + y + z = a + \frac{1}{a}
     \]

3. **Use the convexity of \(f(t)\)**:
   - Since \(f(t) = t + \frac{1}{t}\) is convex for \(t > 0\), by Jensen's inequality:
     \[
     f(y) + f(z) \ge 2f\left(\frac{y+z}{2}\right)
     \]
   - This implies:
     \[
     y + \frac{1}{y} + z + \frac{1}{z} \ge 2\left(\frac{y+z}{2} + \frac{2}{y+z}\right)
     \]
     Simplifying, we get:
     \[
     y + z + \frac{1}{y} + \frac{1}{z} \ge y + z + \frac{4}{y+z}
     \]

4. **Maximize \(x + y + z\)**:
   - We need to maximize \(x + y + z\) under the constraint:
     \[
     \frac{1}{y} + \frac{1}{z} + y + z = a + \frac{1}{a}
     \]
   - Let \(y = z\). Then:
     \[
     2\left(\frac{1}{y} + y\right) = a + \frac{1}{a}
     \]
     Simplifying, we get:
     \[
     2\left(\frac{1}{y} + y\right) = a + \frac{1}{a}
     \]
     Let \(k = y + \frac{1}{y}\). Then:
     \[
     2k = a + \frac{1}{a}
     \]
     Solving for \(k\), we get:
     \[
     k = \frac{a + \frac{1}{a}}{2}
     \]

5. **Evaluate the maximum value**:
   - Since \(y = z\), we have:
     \[
     x + y + z = -a + 2y
     \]
   - Using the constraint \(2\left(y + \frac{1}{y}\right) = a + \frac{1}{a}\), we find:
     \[
     y + \frac{1}{y} = \frac{a + \frac{1}{a}}{2}
     \]
   - The maximum value of \(y + \frac{1}{y}\) occurs when \(y = 1\), giving:
     \[
     y + \frac{1}{y} = 2
     \]
   - Thus, \(a + \frac{1}{a} = 4\), and the maximum value of \(x + y + z\) is:
     \[
     x + y + z = -2 + 2(1) = 0
     \]

The final answer is \(\boxed{0}\).