Consider the sequence $a_0, a_1, a_2, \ldots$ defined by:

$$a_0 = M + \frac{1}{2}$$

and

$$a_{k+1} = a_k \lfloor a_k \rfloor \quad \text{for} \quad k = 0, 1, 2, \ldots$$

We are tasked with finding all positive integers $M$ such that at least one term in the sequence is an integer.

### Analysis of the Sequence

The first term of the sequence is given by:

$$a_0 = M + \frac{1}{2}$$

The floor function $\lfloor a_0 \rfloor$ for $a_0$ is:

$$\lfloor a_0 \rfloor = \lfloor M + \frac{1}{2} \rfloor = M$$

Thus, the sequence proceeds to:

$$a_1 = a_0 \lfloor a_0 \rfloor = \left(M + \frac{1}{2}\right) \times M = M^2 + \frac{M}{2}$$

Then the floor function of $a_1$ is:

$$\lfloor a_1 \rfloor = M^2 + \left\lfloor \frac{M}{2} \right\rfloor$$

Exploring further terms, we see:

$$a_2 = a_1 \lfloor a_1 \rfloor = \left(M^2 + \frac{M}{2}\right) \times (M^2 + \left\lfloor \frac{M}{2} \right\rfloor)$$

### Finding Integer Terms

In order for one of these terms $a_k$ to be an integer, it must be that:

1. For $k=0$, $a_0$ itself must be an integer.
2. For $k \geq 1$, each subsequent term $a_k$ must also be an integer, arising from the multiplication with integer floor values.

For $a_0$ to be an integer, observe:

$$M + \frac{1}{2} \notin \mathbb{Z} \quad \text{since \( M \) is an integer, hence \( a_0 \) is non-integer }$$

For $a_1$ to become an integer, we need $M > 1$. Note:

- If $M = 1$, then $a_0 = 1.5$ and $a_1 = 1.5 \times 1 = 1.5$, which is not integer.
- If $M > 1$, it is possible (though not automatic) that $a_k$ becomes an integer in subsequent iterations due to larger multiplicative products. Particularly, $M \geq 2$ produces $a_1$ values that may drive $a_2$ or further $a_k$ toward integer status under multiplications.

Thus, the condition for having at least one integer term in the sequence is $M > 1$.

Therefore, the solution is:
$$\boxed{M > 1}$$