9.5. For 2011 positive integers, write their $\mathrm{C}_{2011}^{2}$ pairwise sums on the blackboard. Question: Is it possible that exactly half of these $\mathrm{C}_{2011}^{2}$ numbers are multiples of 3, and at the same time, exactly $=$ one-third of them are congruent to 1 modulo 3?
9.5. Let the number of 2011 numbers that are congruent to $0, 1, 2$ modulo 3 be $a, b, c$ respectively. Then
$$
\begin{array}{l}
\frac{a(a-1)}{2}+b c=\frac{b(b-1)}{2}+a c \\
=\frac{c(c-1)}{2}+a b, \\
a+b+c=2011 .
\end{array}
$$

It is not difficult to verify that $a=b=670, c=671$ satisfies the above equations.