To determine whether it is possible to choose $1983$ distinct positive integers, all less than or equal to $10^5$, in such a way that no three of them form a consecutive arithmetic progression, we can approach this problem by constructing a suitable set of integers.

### Step-by-step Construction

1. Understanding the Constraint: We need to ensure that for any selection of three numbers $a, b, c$ from our set, where $a < b < c$, the difference between consecutive terms is not constant, i.e., $b - a \neq c - b$.

2. Strategizing the Selection:
- One common approach is to use numbers that space out sufficiently to prevent forming an arithmetic progression.
- For example, we can choose numbers from the series of triangular numbers or even-numbered positions in arithmetic sequences with large common differences.
- However, a simple and effective strategy involves considering the modulo constraint. Specifically, select numbers such that they do not fall into any simple arithmetic pattern when viewed under a smaller modulus.

3. Utilizing Modular Arithmetic:
- Specifically, we focus on numbers that satisfy a certain congruence condition. For instance, choose numbers of the form $a_k = 1 + 3k$.
- This creates a separation that inherently prevents forming a subset where $b - a = c - b = d$ for some common difference $d$, particularly when $d$ is small compared to the spread achieved by selecting every third number.

4. Selection Feasibility:
- We determine the number of integers that can be chosen using this congruence condition by evaluating the sequence’s reach within the limits.
- For instance, if we select numbers $a_k = 1 + 3k$ up to $10^5$, the number of such numbers is approximately $\left\lfloor \frac{10^5 - 1}{3} \right\rfloor + 1$.

5. Calculation & Conclusion:
- Evaluate the above expression: $\left\lfloor \frac{99999}{3} \right\rfloor + 1 = 33333 + 1 = 33334$.
- This count, $33334$, greatly exceeds 1983, fulfilling our goal of selecting 1983 numbers without forming an arithmetic progression of three terms.

In conclusion, by carefully choosing numbers as described, one can indeed select 1983 distinct positive integers below $10^5$ without forming an arithmetic progression. Thus, the answer is:
$$\boxed{\text{yes}}$$