$4 \cdot 18$ Among four real numbers, if the sum of any three numbers is not less than the other number, then the following statement must be false:
(A) There cannot be only one non-zero number.
(B) All four numbers can be positive.
(C) There are two negative numbers.
(D) If there is a zero, there are no negative numbers.
(China Beijing Junior High School Mathematics Competition, 1990)
[Solution] Let the four real numbers be $a, b, c, d$. According to the problem, we have
$$
\left\{\begin{array}{l}
a+b+c \geqslant d, \\
b+c+d \geqslant a, \\
c+d+a \geqslant b, \\
d+a+b \geqslant c .
\end{array}\right.
$$

If $ad$, from (4) we have $d>c$, which is not appropriate. Similarly, for the other two negative cases, contradictions can also be produced.
Therefore, the answer is $(C)$.