Example 1-19 Select 3 numbers from 1 to 300 so that their sum is exactly divisible by 3. How many schemes are there?
The sum of three numbers is divisible by 3 only in the following three cases:
(1) The three numbers have the same remainder when divided by 3, i.e., all are divisible by 3, or all leave a remainder of 1 or 2.
(2) Among the three numbers, one is divisible by 3, one leaves a remainder of 1 when divided by 3, and one leaves a remainder of 2.

Divide the numbers from $1 \sim 300$ into three groups:
$A=\{1,4,7, \cdots, 298\},$ where the numbers in $A$ leave a remainder of 1 when divided by 3.
$B=\{2,5,8, \cdots, 299\},$ where the numbers in $B$ leave a remainder of 2 when divided by 3.
$C=\{3,6,9, \cdots, 300\},$ where the numbers in $C$ are divisible by 3, i.e., leave a remainder of 0.
There are $C(100,3)$ ways to choose three numbers from each of $A, B, C$.
There are $100^{3}$ ways to choose one number from each of $A, B, C$.
According to the addition principle, the number of ways to choose three numbers such that their sum is divisible by 3 is
$$
\begin{aligned}
3 C(100,3)+100^{3} & =3 \times \frac{100 \times 99 \times 98}{3 \times 2}+1000000 \\
& =1485100
\end{aligned}
$$