Answer: $M=3 n-4$ and $f\left(a_{1}, a_{2}, \ldots, a_{n}\right)=3 n-4$ if and only if $a_{1}=a_{2}=\ldots=a_{n}=3$.

First, we note that for every distribution, there exists a move such that $\max \left(a_{1}, \ldots, a_{n}\right)$ increases by at least one, unless all stones are in one box. Indeed, choose a box where the maximum is achieved, choose another non-empty box, and distribute the stones from that box so that at least one stone goes to the first box. This implies that $f\left(a_{1}, a_{2}, \ldots, a_{n}\right) \leq 3 n-\max \left(a_{1}, \ldots, a_{n}\right)$. In particular, if $\max \left(a_{1}, \ldots, a_{n}\right) \geq 5$, then $f\left(a_{1}, a_{2}, \ldots, a_{n}\right) \leq 3 n-5$. The rest of the proof follows from four claims.

Claim 1. If $\max \left(a_{1}, \ldots, a_{n}\right)=4$, then $f\left(a_{1}, a_{2}, \ldots, a_{n}\right) \leq 3 n-5$.
Proof. Let $A_{1}$ be the box with the most stones, and $A_{2}$ the box with the second-highest number of stones. This means that $a_{1}=4$ and $a_{2} \geq \frac{3 n-4}{n-1}=3-\frac{1}{n-1}$. Since $a_{2}$ is an integer and $n \geq 3$, this means that $a_{2} \geq 3$. We distribute every other box with at least 2 stones so that $A_{1}$ and $A_{2}$ each receive one stone. We repeat this until there are no more boxes with at least 2 stones, except for $A_{1}$ and $A_{2}$. For this distribution $b_{1}, \ldots, b_{n}$, we have $b_{3}+\cdots+b_{n} \leq n-2$. This means that $b_{1}+b_{2} \geq 3 n-(n-2)=2 n+2$ and since $b_{1}-b_{2}=a_{1}-a_{2} \leq 1$, we find that $b_{2}=\frac{1}{2}\left(b_{2}+b_{2}\right) \geq \frac{1}{2}\left(b_{1}+b_{2}-1\right) \geq \frac{1}{2}(2 n+1)=n+\frac{1}{2}$. Since $b_{2}$ is an integer, we find that $b_{2} \geq n+1$. Now we can make a move where we distribute the stones from $A_{2}$, giving $A_{1}$ 2 stones. Then we finish with moves where $A_{1}$ receives at least one stone each time. Since $A_{1}$ receives at least one stone per move and a move involves at least two stones, the number of moves made is at most $3 n-5$.

Claim 2. If $\max \left(a_{1}, \ldots, a_{n}\right)=3$, then $f\left(a_{1}, a_{2}, \ldots, a_{n}\right) \leq 3 n-4$.

Proof. We make an arbitrary move and apply Claim 1 to the result. Then we are done in at most $1+(3 n-5)=3 n-4$ moves. Alternatively, we can give a concrete series of $3 n-4$ moves. If $\max \left(a_{1}, \ldots, a_{n}\right)=3$, each box has exactly 3 stones. We start with $n-2$ moves where we choose $A_{i}$ with $i \geq 3$ and distribute the three stones over $A_{1}, A_{2}$, and $A_{i}$. Then the first two boxes each have $n+1$ stones, so we can make a move where we distribute the $n+1$ stones from the second box so that the first box gets 2 stones. Since $A_{1}$ receives at least one stone per move and a move involves at least two stones, the number of moves made is at most $3 n-4$.

Claim 3. There are no moves such that the maximum $\max \left(a_{1}, \ldots, a_{n}\right)$ increases by 3 or more.

Proof. To make a move where a box receives more than 3 stones, the chosen box would need to contain at least $3 n+1$ stones. This contradicts the fact that there are only $3 n$ stones. To make a move where a box receives 3 stones, the chosen box must contain at least $2 n+1$ stones. A box that achieves the new maximum after this move must also contain at least $2 n+1$ stones. This is again a contradiction.

Claim 4. If $\max \left(a_{1}, \ldots, a_{n}\right)=3$, then $f\left(a_{1}, a_{2}, \ldots, a_{n}\right) \geq 3 n-4$.
Proof. Suppose we can get all the stones into one box in $3 n-5$ or fewer turns. By Claim 3, there are no moves where the maximum increases by 3 or more. This means that there are at least two moves where the maximum increases by 2. We call such moves large moves. Let the first large move be in turn $i$ and the last large move be in turn $j$. Since a large move can only be performed with a box containing at least $n+1$ stones and each box starts with 3 stones, we have $i-1 \geq(n+1)-3$ or $i \geq n-1$.

Let $m$ be the number of empty boxes at the start of move $j$. Since each box has at least 1 stone after move $i$, we must have made at least one move per empty box after that. This means that $(j-1)-i \geq m$. Since each box receives at least 1 stone, there are at most $m$ boxes with 1 stone after move $j$; the rest contain at least 2. Since we do not make any large moves after move $j$, it takes at least $m+2(n-1-m)=2(n-1)-m$ moves to empty $n-1$ boxes. The total number of moves is thus at least

$$i+(j-i)+2(n-1)-m \geq(n-1)+(m+1)+2(n-1)-m=3 n-2,$$

which contradicts our assumption that we could do it in $3 n-5$ moves.