4. Find all non-zero tuples $(x, y, z, w)$ that satisfy
$$
\begin{array}{l}
1+\frac{1}{x}+\frac{2(x+1)}{x y}+\frac{3(x+1)(y+2)}{x y z}+ \\
\frac{4(x+1)(y+2)(z+3)}{x y z w}=0
\end{array}
$$
4. Notice that, the sum of the first three terms in the original expression is
$$
\left(1+\frac{1}{x}\right)\left(1+\frac{2}{y}\right),
$$

adding the fourth term gives
$$
\left(1+\frac{1}{x}\right)\left(1+\frac{2}{y}\right)\left(1+\frac{3}{z}\right),
$$

and adding the fifth term, the original expression is equivalent to
$$
\left(1+\frac{1}{x}\right)\left(1+\frac{2}{y}\right)\left(1+\frac{3}{z}\right)\left(1+\frac{4}{w}\right)=0 .
$$

Thus, $1+\frac{1}{x}=0$ or $1+\frac{2}{y}=0$ or $1+\frac{3}{z}=0$ or $1+\frac{4}{w}=0$, the solution set is the set of $(x, y, z, w)$.
Since $x y z w \neq 0$, we have $x=-1$ or $y=-2$ or $z=-3$ or $w=-4$.