4. The equation
$$
x y z+x y+y z+z x+x+y+z=2014
$$

has $\qquad$ groups of non-negative integer solutions $(x, y, z)$.
4. 27 .

The original equation can be transformed into $(x+1)(y+1)(z+1)=2015=5 \times 13 \times 31$.
If $x, y, z$ are all positive integers, then there are $1 \times 2 \times 3=6$ solutions; $\square$

If one of $x, y, z$ is zero and the other two are positive integers, then there are $3 \times 6=18$ solutions;

If two of $x, y, z$ are zero and one is a positive integer, then there are 3 solutions;
Therefore, the original equation has $6+18+3=27$ non-negative integer solutions.