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Let, $k_i$ be the stone on which ith frog was initially sitting on. Let's say all move M times. Then each frog will naturally go from $k_i \to k_i +M$(Mod 23). We can work on (Mod 23)

Now if 2 or more frogs sit on the same stone then we will call one(any) of them 'domain' and the rest 'refugees'. (I know I am bad at making names). It will be sufficient if we prove that at some point there will be at least 5 refugees.

Now, the value of M(the number of moves) is between (1-23) as after 23 moves all will again come at the same initial position. So we only need to think about (1-23) values of M.

Claim-1: Each pair of frogs will have a value M(1-23) such that after M moves that pair will be on the same stone.
Proof:
To make our claim true the congruence must hold in (mod 23) of course.
$k_i+iM \equiv k_j +jM$
Or,$k_i - k_j \equiv (j-i)M$
The second form holds as $(k_i - k_j) \neq 0$ and is in the complete residue class of $23$ and so is $j-i$. So we can find a value M for which the congruence holds. And M will also be in (1-23). Note this holds only because 23 is prime. $\square$

Now, there are $\binom{22}{2}=231$ pairs of frogs. And each pair will have a value M. Since there are 23 different possible values of M(or at least the ones we need to look at, by PHP we get, $\left\lceil \frac{231}{23} \right\rceil = 11$. So 11 pairs must have the same value of M.

Claim-2: There will be at least 5 refugees at some point
Proof:
Let's say after $M_1$ moves [Let $M_1$ be the value of the 11 pairs] 11 of the pairs will sit on the same stone. If there were less than 5 refugees at that time we cannot find 11 pairs who sit on the same stone. A contradiction. So there must be at least 5 refugees at that time.$\square$

So there will be at least 6 stones empty at some point. $\square$