Solution:

Denote by $L_{i}, 0 \leq i < n$, the number of tiles that John puts in such a way that $a_{0}$ is added at position $i$. Note that $L_{n-k} = L_{n-k+1} = \cdots = L_{n-1} = 0$. Analogously, we define $R_{i}, 0 \leq i < n$ for the number of times $a_{k-1}$ was added at position $i$.

First, note that $a_{0} L_{0} = K$, since only the leftmost operations add something to the first zero. Note that for all $0 \leq i < k$ we have $K \geq a_{i} L_{0} = a_{i} K / a_{0}$, hence $a_{i} \leq a_{0}$. Analogously, focusing on $a_{k-1}$ and $R_{n-1}$ we conclude that for every $0 \leq i < k$ we have $a_{i} \leq a_{k-1}$. That is, we have $a_{0} = a_{k-1}$ and this is the largest value in the sequence $a_{0}, \ldots, a_{k-1}$; moreover, $L_{0} = R_{n-1}$. To simplify notation, we denote $a = a_{0} = a_{k-1}$ and $p = L_{0} = R_{n-1}$.

Our task is to prove that every $a_{i}$ has value either $0$ or $a$. We will prove that by induction, in fact proving the following, stronger, statement. For every $0 \leq i < \lceil k / 2 \rceil$, we prove:

(a) $a_{i} = a_{k-1-i}$,

(b) $L_{i} = R_{n-1-i}$,

(c) either $a_{i} = 0$ or $a_{i} = a$.

In plain words, besides proving the required statement, we are also gradually proving that the sequence $a_{i}$ is symmetrical, as well as John's moves.

We have already proven this statement for $i = 0$. Let us now prove it for any $i$, assuming it was proven for $0, \ldots, i-1$.

Note that for the $i$-th position on the blackboard we have
$$K = \sum_{j=0}^{i} L_{j} a_{i-j}$$
Let us first focus on the terms $L_{1} a_{i-1}, L_{2} a_{i-2}, \ldots, L_{i-1} a_{1}$. By the induction statement, the value of each of these terms is either $0$ or $p \cdot a = K$. If the latter ever happens, the induction statement implies $L_{i} = R_{n-1-i} = 0$ together with $a_{i} = a_{k-1-i} = 0$ and we are finished.

Hence, we can next assume that
$$K = L_{0} a_{i} + L_{i} a = p a_{i} + L_{i} a$$
and the analogous argument on the other side (replace $L_{j}$ by $R_{n-1-j}$ and $a_{j}$ by $a_{k-1-j}$ ) allows us to assume that
$$K = R_{n-1} a_{k-1-i} + R_{n-1-i} a$$

Consider the case $a_{i} < a$. Equation (1) implies that $L_{i} > 0$. Focusing now at the $k$th number in John's sequence, we get $K = L_{0} a_{k-1} + \cdots + L_{i} a_{k-1-i} + \ldots$. But $L_{0} a_{k-1} = p a = K$, so we are getting $L_{i} a_{k-1-i} = 0$, hence $a_{k-1-i} = 0$.

We can apply exactly the same argument on the other side (replace $L_{j}$ with $R_{n-1-j}$ and $a_{j}$ with $a_{k-1-j}$ and eq. (1) with eq. (2)). We get that whenever $a_{k-1-i} < a$, it has to be that $a_{i} = 0$.

Putting these two arguments together, we conclude that one of the two possibilities is true: Either $a_{i} = a_{k-1-i} = a$ and eqs. (1) and (2) imply $L_{i} = R_{n-i} = 0$. Or $a_{i} = a_{k-1-i} = 0$ and eqs. (1) and (2) imply $L_{i} = R_{n-i} = p$. In either case, the induction statement is proven for $i$ and we are done.

Number-theoretic solution. As in the first solution, we denote by $L_{i}$ the number of times that $a_{0}$ was added to the $i$-th position and note that $a_{0}$ is maximal among the terms of the sequence. Obviously, we are done if $a_{0} = 0$, so let's assume $a_{0} > 0$, implying $L_{0} > 0$. We normalize the sequence by dividing each term by $a_{0}$, so that $a_{0} = 1$ and all numbers on the board are equal to $L_{0}$ in the end.

Claim. All terms of the sequence are now rational numbers.

Proof. Assume the contrary. Consider the smallest index $i$ with $a_{i} \notin \mathbb{Q}$ and look at the $i$-th number on the board which in the end of the process takes the value $L_{0} \in \mathbb{N}$. Since $a_{i}$ was added to this position $L_{0}$ times and all other terms added to it must have been rational by minimality of $i$, we get a contradiction.

If there now exists some index $j$ with $0 < a_{j} < 1$, write $a_{j}$ as a reduced fraction and take a prime factor $q$ of its denominator. In other words, $v_{q}\left(a_{j}\right) < 0 = v_{q}\left(a_{0}\right)$.

We now choose a minimal index $t$ that minimizes $v_{q}\left(a_{t}\right)$ and a minimal index $s$ that minimizes $v_{q}\left(L_{s}\right)$. By evaluating the number on the $(s+t)$-th position on the blackboard, we obtain
$$L_{0} \cdot a_{0} = \sum_{i=0}^{s+t} L_{i} \cdot a_{s+t-i}$$
Here, we simply set $a_{s+t-i} = 0$ if $s+t-i \geq k$. Note that the term $L_{s} \cdot a_{t}$ appearing in this sum has strictly smaller $v_{q}$ than any other term in the sum, by definition and minimality of $s$ and $t$. Also, since $v_{q}\left(L_{s}\right) \leq v_{q}\left(L_{0}\right)$ and $v_{q}\left(a_{t}\right) < v_{q}\left(a_{0}\right) = 0$, we have $v_{q}\left(L_{0} \cdot a_{0}\right) > v_{q}\left(L_{s} \cdot a_{t}\right)$. Hence,
$$v_{q}\left(L_{0} \cdot a_{0}\right) > v_{q}\left(L_{s} \cdot a_{t}\right) = v_{q}\left(\sum_{i=0}^{s+t} L_{i} \cdot a_{s+t-i}\right)$$
contradiction. We conclude that no such $j$ could have existed, so all terms are either equal to $0$ or $a_{0}$.

Solution using Polynomials. By the claim of the previous solution, we see that we may assume all terms of the sequence to be nonnegative integers (just multiply everything by the lcm of the denominators). Again, we note that for all $i$, it holds that $a_{0} \geq a_{i}$.

Now observe the polynomial
$$P(x) = \sum_{i=0}^{k-1} a_{i} x^{i}$$
and interpret the numbers on the board as a polynomial in a similar way, i.e. if the numbers on the board are $b_{0}, \ldots, b_{n-1}$, read it as
$$\sum_{i=0}^{n-1} b_{i} x^{i}$$
The problem states that after each step, the polynomial on the board is increased by $x^{p} \cdot P(x)$ for some $p \in \mathbb{N}_{0}$. Therefore, the condition can be rewritten as
$$P(x) \cdot Q(x) = K \cdot \frac{x^{n} - 1}{x - 1}$$
for some polynomial $Q \in \mathbb{Z}[x]$ and $K \in \mathbb{N}$. It follows that
$$P(x) = a \cdot R(x)$$
where $a \mid K$ and $R(x) \left\lvert\, \frac{x^{n} - 1}{x - 1}\right.$. We see that the constant term of $R$ must be equal to $1$, and so $a_{0} = a$. As $a \mid a_{i}$ and $a_{0} \geq a_{i}$ for all $i$, they are indeed all equal to $a$.