Solution:

We have $a_{0}=m+\frac{1}{2}$ and $a_{1}=a_{0}\left\lfloor a_{0}\right\rfloor=\left(m+\frac{1}{2}\right) \cdot m=m^{2}+\frac{m}{2}$. This expression is obviously an integer for even $m$, so that the desired property of the sequence holds in this case.

Furthermore, for $m=1$ we have $a_{0}=\frac{3}{2}$, $\left\lfloor a_{0}\right\rfloor=1$, and for $a_{k}=\frac{3}{2}$ we always also have $a_{k+1}=\frac{3}{2} \cdot 1=\frac{3}{2}$. Here, then, there is no integer term of the sequence.

Now let $m \geq 3$ be odd. There exists the unique representation $m=2^{p} \cdot n_{0}+1$ with $p \geq 1$, $n_{0}$ odd and $p, n_{0} \in \mathbb{N}$. Thus $a_{0}=2^{p} n_{0}+\frac{3}{2}$, and for $a_{k}=2^{p} n_{k}+\frac{3}{2}$ ($n_{k}$ odd) it follows that
$$a_{k+1}=\left(2^{p} n_{k}+\frac{3}{2}\right)\left(2^{p} n_{k}+1\right)=2^{p-1}\left(2^{p+1} n_{k}^{2}+5 n_{k}\right)+\frac{3}{2},$$
where the term in parentheses is odd. Hence there exists a representation $a_{k+1}=2^{p-1} n_{k+1}+\frac{3}{2}$ with $n_{k+1}$ odd. Thus, as $k$ is increased by $1$, $p$ is simultaneously decreased by $1$, so that $a_{k+p}=2^{0} n_{k+p}+\frac{3}{2}$ holds and $\left\lfloor a_{k+p}\right\rfloor$ is even. Then $a_{k+p+1}$ is an integer, and thus the desired property of the sequence also holds for every $m \geq 3$.

Thus all positive integers except $1$ belong to the sought set.