Example 3 Can we select 1983 different integers from the non-negative integers less than $10^{5}$, such that no three of them form an arithmetic progression?
【Analysis】It can be done.
First, each time "greedily" take the smallest number that does not form an arithmetic sequence with the existing terms, and enumerate to get
$$
0,1,3,4,9,10,12,13,27,28, \cdots
$$

When written in ternary, they are sequentially
$$
0,1,10,11,100,101,110,111,1000,1001, \cdots
$$

Observing, we see that only 0 or 1 appears.
The binary representation of 1982 is 11110111110. Converting this number from ternary to decimal, we find that the 1983rd number is $87843<100000$.

Next, we prove that if every term in the sequence has a ternary representation consisting only of 0 or 1, then it is impossible for any three terms to form an arithmetic sequence. Otherwise, if three terms $x<y<z$ form an arithmetic sequence, i.e., $z=2 y-x$, since
$$
0 \times 2-1=2,1 \times 2-0=2 \text {, }
$$

then the difference at the rightmost differing digit between $x$ and $y$ is 2, which is a contradiction.