4. (6 points) Non-zero digits $a, b, c$ can form 6 different three-digit numbers without repeated digits, and the sum of these 6 numbers is 5994. Then any one of these 6 numbers can $\qquad$ be divided by 9. (Fill in “can” or “cannot”)
【Solution】Solution: According to the analysis, these six numbers are: $\overline{\mathrm{abc}}, \overline{\mathrm{acb}}, \overline{\mathrm{bac}}, \overline{\mathrm{bca}}, \overline{\mathrm{cab}}, \overline{\mathrm{cba}}$, according to the principle of place value:
$$
\begin{array}{l}
\overline{a b c}=100 a+10 b+c \quad \overline{a c b}=100 a+10 c+b \\
\overline{b a c}=100 b+10 a+c \quad \overline{b c a}=100 b+10 c+a \\
\overline{c a b}=100 \mathrm{c}+10 \mathrm{a}+\mathrm{b} \quad \overline{\mathrm{cba}}=100 \mathrm{c}+10 \mathrm{~b}+\mathrm{a} \\
\overline{a b c}+\overline{a c b}+\overline{b a c}+\overline{b c a}+\overline{c a b}+\overline{c b a} \\
=100 a+10 b+c+100 a+10 c+b+100 b+10 a+c+100 b+10 c+a+100 c+10 a+b+100 c+10 b+a \\
=200(a+b+c)+20(a+b+c)+2(a+b+c) \\
=222(a+b+c) \\
=5994 \\
\therefore a+b+c=27 \\
\because 27 \div 9=3 \\
\end{array}
$$
$\therefore$ The characteristic of being divisible by 9 is that the sum of the digits can be divided by 9, and the sum of the digits of any one of these six numbers is: $a+b+c=27$. It is easy to see that any of these 6 numbers can be divided by 9. Therefore, the answer is: Yes