The required integers are $p-1$ along with all the numbers of the form $\lfloor p / q\rfloor, q=$ $2, \ldots, p-1$. In other words, these are $p-1$, along with the numbers $1,2, \ldots,\lfloor\sqrt{p}\rfloor$, and also the (distinct) numbers $\lfloor p / q\rfloor, q=2, \ldots,\left\lfloor\sqrt{p}-\frac{1}{2}\right\rfloor$.

We begin by showing that these numbers satisfy the conditions in the statement. It is readily checked that $p-1$ satisfies the required inequalities, since $m+R(m(p-1))=m+(p-m)=$ $p>p-1$ for all $m=1, \ldots, p-1$.

Now, consider any number $a$ of the form $a=\lfloor p / q\rfloor$, where $q$ is an integer greater than 1 but less than $p$; then $p=a q+r$ with $0 \leq r < a$. Thus, $a$ satisfies the required condition. If $m \leq q$, then $m a \leq q a = p - r < p$, so $R(m a) = m a$ and $m + R(m a) = m + m a = m (a + 1) \leq q (a + 1) = p + q - r < p$. If $m > q$, then $m a \geq (q + 1) a = p + a - r > p$, so $R(m a) = m a - p = m a - (a q + r) = m a - a q - r = a (m - q) - r$ and $m + R(m a) = m + a (m - q) - r = m + a m - a q - r = m (a + 1) - a q - r = m (a + 1) - p + r > p$.

Finally, we show that if an integer $a \in(0, p-1)$ satisfies the required condition then $a$ is indeed of the form $a=\lfloor p / q\rfloor$ for some integer $q \in(0, p)$. This is clear for $a=1$, so we may (and will) assume that $a \geq 2$.

Write $p=a q+r$ with $q, r \in \mathbb{Z}$ and $0 \leq r < a$; since $a \geq 2$ we have $q < p / 2$. Choose $m=q+1<p$; we have $R(m a)=R(a q+a)=R(p+(a-r))=a-r$, so

$$a<m+R(m a)=q+1+a-r,$$

which yields $r<q+1$. Moreover, if $r=q$, then $p=q(a+1)$ which is impossible by $1<a+1<p$. Thus $r<q$, and we have

$$0 \leq \frac{p}{q}-a=\frac{r}{q}<1,$$

which proves $a=\lfloor p / q\rfloor$.