N1. Determine all positive integers $M$ for which the sequence $a_{0}, a_{1}, a_{2}, \ldots$, defined by $a_{0}=\frac{2 M+1}{2}$ and $a_{k+1}-a_{k}\left\lfloor a_{k}\right\rfloor$ for $k-0,1,2, \ldots$, contains at least one integer term.
Answer. All integers $M \geqslant 2$.
Solution 1. Define $b_{k}-2 a_{k}$ for all $k \geqslant 0$. Then
$$
b_{k+1}=2 a_{k+1}-2 a_{k}\left\lfloor a_{k}\right]-b_{k}\left\lfloor\frac{b_{k}}{2}\right\rfloor .
$$

Since $b_{0}$ is an integer, it follows that $b_{k}$ is an integer for all $k \geqslant 0$.
Suppose that the sequence $a_{0}, a_{1}, a_{2}, \ldots$ does not contain any integer term. Then $b_{k}$ must be an odd integer for all $k \geqslant 0$, so that
$$
b_{k+1}=b_{k}\left\lfloor\frac{b_{k}}{2}\right\rfloor-\frac{b_{k}\left(b_{k}-1\right)}{2} .
$$

Hence
$$
b_{k+1}-3=\frac{b_{k}\left(b_{k}-1\right)}{2}-3-\frac{\left(b_{k}-3\right)\left(b_{k}+2\right)}{2}
$$
for all $k \geqslant 0$.
Suppose that $b_{0}-3>0$. Then equation (2) yields $b_{k}-3>0$ for all $k \geqslant 0$. For each $k \geqslant 0$, define $c_{k}$ to be the highest power of 2 that divides $b_{k}-3$. Since $b_{k}-3$ is even for all $k \geqslant 0$, the number $c_{k}$ is positive for every $k \geqslant 0$.

Note that $b_{k}+2$ is an odd integer. Therefore, from equation (2), we have that $c_{k+1}-c_{k}-1$. Thus, the sequence $c_{0}, c_{1}, c_{2}, \ldots$ of positive integers is strictly decreasing, a contradiction. So, $b_{0}-3 \leqslant 0$, which implies $M=1$.

For $M=1$, we can check that the sequence is constant with $a_{k}-\frac{3}{2}$ for all $k \geqslant 0$. Therefore, the answer is $M \geqslant 2$.

Solution 2. We provide an alternative way to show $M-1$ once equation (1) has been reached. We claim that $b_{k}=3\left(\bmod 2^{m}\right)$ for all $k \geqslant 0$ and $m \geqslant 1$. If this is true, then we would have $b_{k}-3$ for all $k \geqslant 0$ and hence $M=1$.

To establish our claim, we proceed by induction on $m$. The base case $b_{k}=3(\bmod 2)$ is true for all $k \geqslant 0$ since $b_{k}$ is odd. Now suppose that $b_{k}=3\left(\bmod 2^{m}\right)$ for all $k \geqslant 0$. Hence $b_{k}=2^{m} d_{k}+3$ for some integer $d_{k}$. We have
$$
3=b_{k+1}=\left(2^{m} d_{k}+3\right)\left(2^{m-1} d_{k}+1\right)=3 \cdot 2^{m-1} d_{k}+3 \quad\left(\bmod 2^{m}\right),
$$
so that $d_{k}$ must be even. This implies that $b_{k}=3\left(\bmod 2^{m+1}\right)$, as required.
Comment. The reason the number 3 which appears in both solutions is important, is that it is a nontrivial fixed point of the recurrence relation for $b_{k}$.