We claim the max number of toads that the princess can place is $\left\lceil \frac{n}{2} \right\rceil.$ Let $a_1, a_2, \dots, a_n$ the labels of the rock in order ( after $a_n$ comes $a_1$ again). We consider $a_{n+i}=a_i$ for the rest of the proof. Claim: If there's a toad on $a_l$ and a toad on $a_k$, they are never going to meet if and only if $a_l+a_{l+1}+\dots+a_{k-1}\ge a_m$ for $1\le m\le n.$ Proof: Notice that if $a_l+a_{l+1}+\dots+a_{k-1}<a_k$, the two toads are going to meet on $a_k$ at the minute $a_l+a_{l+1}+\dots+a_{k-1}$, then $a_l+a_{l+1}+\dots+a_{k-1}\ge a_k$. Now for any $q\ge k$, if $a_l+a_{l+1}+\dots+a_{k-1}+a_k+ a_{k+1}+\dots +a_{q-1}<a_k+ a_{k+1}+\dots +a_{q}$, the two toads are going to meet on $a_q$ at the minute $a_l+a_{l+1}+\dots+a_{q-1}$, then $$a_l+a_{l+1}+\dots +a_{q-1}\ge a_k+ a_{k+1}+\dots +a_{q} \Rightarrow a_l+a_{l+1}+\dots +a_{k-1}\ge a_{q}.$$ Now back to the problem lets consider the princess labels the rock in groups of two that add $n$ the following way: $$n, 1, n-1, 2, n-2, \dots k, n-k, \dots$$If the princess places a toad on the first of each group, meaning on $n$, and on each $k$ with $k<n/2$ this configuration works and has $\left\lceil \frac{n}{2} \right\rceil$ toads. If $n$ is even she places $n/2$ before $n$ and the configuration still works. This is because the distance between any two toads is at least 2 rocks (except for the case of the frog on $n$) and thanks to the configuration the sum of the labels of this two rocks is $n$ and then using the lemma we conclude that no two toads are going to meet. To finish the problem consider that the princess places $ \left\lceil \frac{n}{2} \right\rceil+1$ toads. By the pigeonhole principle this means there's at least two toads with a toad directly in front of them. Meaning there's a $k$ such that there's a toad on $a_k$ and a toad on $a_{k+1}$ and $a_k<n$. Using the lemma we can conclude that this two toads we'll meet on the rock with the label $n$ and hence is imposible for the princess to place this many toads.

The maximum number of toads that the princess can place is $\left \lfloor \frac{n+1}{2}\right \rfloor$. We change the process in the following way: we replace a stone labeled with the number $k$ by $k$ stones labeled with the number $k$, as such the stones form blocks of $k$ consecutives stones with label $k$. For example we would change the configuration \[1423 \text{ to the configuration }1444422333\]Now each toad jumps to the next stone every minute and we want that no two toads stand on stones with the same label at the same time. We also have to consider that in a valid starting configuration the toads have to be on the first stone of every block. It is easy to see that this is equivalent to the original process. We say that the distance between two toads is the smallest number of stones that one would have to jump to get from one toad to the other one, note that the distance between toads remains constant as the process runs. As there are $n$ consecutive stones labeled $n$, if the distance between two toads is less than $n$ then at some point the two toads will be both on stones labeled $n$ which is imposible. Thus the distance between any two toads is at least $n$. As the total number of stones is $1+2+\dots +n=\frac{n(n+1)}{2}$ and every two toads are at distance at least $n$, then there are at most $\frac{n(n+1)}{2n}=\frac{n+1}{2}$ toads which gives us our desired bound. For brevity we will write $k$ to denote the block of $k$ stones labeled $k$.To see that $\left \lfloor \frac{n+1}{2}\right \rfloor$ is attainable the princess first puts the blocks in the following way: \[n,1,n-1,2,n-2,3,n-3,\dots ,k,n-k,\dots\]and then she puts the toads on the first stones of the blocks $n,1,2,3,\dots,\left \lfloor \frac{n+1}{2}\right \rfloor-1$. By doing so, the distance between any two toads is at least $n$, and thus it is imposible for two toads to be on the same block at the same time as all blocks have a length smaller than or equal to $n$.