Let's find the integers $x, y, z$ such that

$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}
$$
The fractions in the equation are meaningful when $x y z(x+y+z) \neq 0$. After multiplying by the denominators,

$$
(x+y+z)(x y+y z+z x)=x y z
$$

is obtained.

After expanding and rearranging according to the powers of $x$:

$$
x^{2}(y+z)+x(y+z)^{2}+y z(y+z)=0
$$

After factoring out $(y+z)$ on the left side,

$$
(y+z)\left(x^{2}+x(y+z)+y z\right)=(y+z)(x+z)(x+y)
$$

Therefore, the solutions to the equation are precisely those triples of numbers where none of the elements are zero and among the three elements, two are opposites of each other.