31. From the 100 numbers $1,2,3, \cdots, 100$, if 3 numbers are randomly selected, find the probability that their sum is divisible by 3.
31. First, classify these 100 numbers based on their remainders when divided by 3.

By
$$
\begin{array}{l}
1 \leqslant 3 k \leqslant 100 \Rightarrow \frac{1}{3} \leqslant k \leqslant 33 \frac{1}{3}, \\
1 \leqslant 3 k+1 \leqslant 100 \Rightarrow 0 \leqslant k \leqslant 33 .
\end{array}
$$

Thus, there are 33 numbers of the form $3 k$, 34 numbers of the form $3 k+1$, and 33 numbers of the form $3 k+2$. These are denoted as $A$ type, $B$ type, and $C$ type, respectively. The numbers that meet the criteria can be divided into 4 categories:
(1) 3 $A$ type; (2) 3 $B$ type; (3) 3 $C$ type; (4) 1 $A$ type, 1 $B$ type, 1 $C$ type.
Therefore, the required probability is
$$
P=\frac{C_{33}^{3}+C_{34}^{3}+C_{33}^{3}+C_{33}^{1} \cdot C_{34}^{1} \cdot C_{33}^{1}}{C_{100}^{3}}=\frac{817}{2450} .
$$