1. Let $\lfloor x\rfloor$ denote the greatest integer not exceeding the real number $x$. Find all positive integers $M$ such that the sequence defined by
$$
\begin{array}{l}
a_{0}=\frac{2 M+1}{2}, \\
a_{k+1}=a_{k}\left\lfloor a_{k}\right\rfloor(k=0,1, \cdots)
\end{array}
$$

contains at least one integer term.
1. $M$ is the set of all integers greater than or equal to 2.

For all non-negative integers $k$, define $b_{k}=2 a_{k}$, then $b_{k+1}=2 a_{k+1}=2 a_{k}\left\lfloor a_{k}\right\rfloor=b_{k}\left\lfloor\frac{b_{k}}{2}\right\rfloor$.

Since $b_{0}$ is an integer, thus, for all non-negative integers $k, b_{k}$ are all integers.

Assume the sequence $a_{0}, a_{1}, \cdots$ does not contain any integer terms. Then for all non-negative integers $k, b_{k}$ are all odd.
Hence $b_{k+1}=b_{k}\left\lfloor\frac{b_{k}}{2}\right\rfloor=\frac{b_{k}\left(b_{k}-1\right)}{2}$.
Thus, for all non-negative integers $k$, we have
$$
\begin{array}{l}
b_{k+1}-3=\frac{b_{k}\left(b_{k}-1\right)}{2}-3 \\
=\frac{\left(b_{k}-3\right)\left(b_{k}+2\right)}{2} .
\end{array}
$$

If $b_{0}-3>0$, by equation (1), we know that for all non-negative integers $k$, we have $b_{k}-3>0$.

For each non-negative integer $k$, define the non-negative integer $c_{k}$ such that $2^{c_{k}} \|\left(b_{k}-3\right)$.

Since for all non-negative integers $k$, $b_{k}-3$ is even, thus, $c_{k}$ is a positive integer.
Also, $b_{k}+2$ is odd, by equation (1) we know
$c_{k+1}=c_{k}-1$.
Thus the positive integer sequence $c_{0}, c_{1}, \cdots$ is strictly decreasing, which is a contradiction.
Therefore, $b_{0}-3 \leqslant 0$. This indicates that $M=1$.
If $M=1$, then for all non-negative integers $k$, we have $a_{k}=\frac{3}{2}$, which does not satisfy the problem's condition.
Therefore, the positive integer $M \geqslant 2$.