Answer: The only such sequences are $a_{n}=n+C$, where $C$ is a nonnegative integer.

## Common remarks.
- Denote by $a_{[i, j]}$ the subsequence $a_{i}, a_{i+1}, \ldots, a_{j}$.
- Without loss of generality, in each solution we suppose $p<q$. It can be convenient to treat the case where $p=1$ separately.
- The problem can also be posed for sequences of arbitrary integers (rather than positive). Refer to the comment after Solution 1 for a proof.

Solution 1. Let $k=\left\lceil\frac{q}{p}\right\rceil$. Note that $k \geqslant 2$.

Lemma 1. If $i, j$ and $m$ are positive integers such that $|i-j| \leqslant m p$ then $\left|a_{i}-a_{j}\right| \leqslant m p$.

Proof. By the given condition, if $|i-j| \leqslant p$ then $\left|a_{i}-a_{j}\right| \leqslant p$. So the lemma follows from induction on $m$ and the triangle inequality.

Lemma 2. For a fixed $n$, suppose that $a_{i}$ is minimal over $i \geqslant n$. Then $i \leqslant n+p-1$.

Proof. Suppose for contradiction that $i \geqslant n+p$. Then $\min \left(a_{[i-p, i+q-p]}\right)=a_{i}$. Since $q-p \leqslant (k-1) p$, it follows from Lemma 1 that $\max \left(a_{[i-p, i+q-p]}\right) \leqslant a_{i}+(k-1) p<a_{i}+q$, which is a contradiction.

Lemma 3. For a fixed $n>q$, suppose that $a_{i}$ is maximal over $i \leqslant n$. Then $i \geqslant n-p+1$.

Proof. Suppose $a_{j}$ is minimal over $j \geqslant n-q$. Then by Lemma $2, j \leqslant n-q+p-1$. So $\min \left(a_{[n-q, n]}\right)=a_{j}$ and $a_{i} \geqslant \max \left(a_{[n-q, n]}\right)$, which implies that $a_{i} \geqslant a_{j}+q$.

Lemma 2 also implies that if $j \geqslant n$ then $a_{j} \geqslant \min \left(a_{[n, n+p]}\right)$. So if $i<j$, then we have $a_{j} \geqslant a_{i}-p$, which contradicts $a_{i} \geqslant a_{j}+q$. Hence we must have $i>j$.

The above inequality also gives $\left|a_{i}-a_{j}\right| \geqslant q>(k-1) p$, so by Lemma 1 it follows that $|i-j|>(k-1) p$. Therefore $i>j+(k-1) p \geqslant n-q+(k-1) p \geqslant n-p+1$.

Let $b_{n}$ be the minimal value of $a_{i}$ for $i \geqslant n$. By Lemma 2, $b_{n+p}>b_{n}$ for all $n$. Hence $b_{n}=\min \left(a_{[n, n+p]}\right)=\min \left(a_{[n, n+q]}\right)$. Let $c_{n}$ be the maximal value of $a_{i}$ for $i \leqslant n$. By Lemma 3, $c_{n-p}>c_{n}$ for all $n>q$. Hence $c_{n}=\max \left(a_{[n-p, n]}\right)=\max \left(a_{[n-q, n]}\right)$ for $n>q$.

So if $n>q$ then $b_{n}=c_{n+p}-p=c_{n+q}-q$. So for $n>q$ we get $b_{n+q-p}+p=c_{n+q}=b_{n}+q$, and hence $b_{n+q-p}=b_{n}+q-p$.

Next note that $b_{n+p} \leqslant a_{n+p} \leqslant b_{n}+p$. So $b_{n+p}-b_{n} \leqslant p$ for all $n>q$, and iterating this ($q-p$) times gives $b_{n+p(q-p)}-b_{n} \leqslant p(q-p)$. But using $b_{n+q-p}=b_{n}+q-p$ gives $b_{n+p(q-p)}-b_{n}=p(q-p)$. Since equality occurs, we must have $b_{n+p}=b_{n}+p$.

So for $n>q, b_{n+p}=b_{n}+p$ and $b_{n+q-p}=b_{n}+q-p$. Since $p$ and $q-p$ are coprime, $b_{n+1}=b_{n}+1$ for all $n>q$. The only way for $b_{n}$ and $b_{n+1}$ to be different is if $b_{n}=a_{n}$, so we deduce that $a_{n+1}=a_{n}+1$ and there is a constant $C$ such that $a_{n}=n+C$ for all $n>q$.

Finally, suppose $a_{n}=n+C$ for all $n \geqslant N$. Then $p=\max \left(a_{N-1}, N+C+p-1\right)- \min \left(a_{N-1}, N+C\right)$. So $a_{N-1}=N+C+p$ or $N+C-1$. Similarly, $a_{N-1}=N+C+q$ or $N+C-1$. Hence $a_{N-1}=N+C-1$. So, by induction, we have $a_{n}=n+C$ for all positive integers $n$. Since $a_{1} \geqslant 1, C$ is a nonnegative integer.

It is trivial to check that $a_{n}=n+C$ satisfies the given condition.

For $n, x \geqslant 1$, let the $x$-width of $n$ be $\max \left(a_{[n, n+x]}\right)-\min \left(a_{[n, n+x]}\right)$. We call a positive integer $x$ good if the $x$-width of $n$ is less than or equal to $x$ for all sufficiently large $n$, and we call $x$ very good if the $x$-width of $n$ is equal to $x$ for sufficiently large $n$.

Lemma 1. If $p'$ is good and $q'$ is very good with $p'<q'<2p'$, then $2p'-q'$ is also good.

Proof. Note that $0<q'-p'<p'<q'$. Let $n$ be a sufficiently large positive integer. Then for $k \in [n+q'-p', n+p']$, we have $a_{k} \geqslant \max \left(a_{[n, n+p']}\right)-p'$ and $a_{k} \geqslant \max \left(a_{[n+q'-p', n+q']}\right)-p'$ since $p'$ is good, which shows $a_{k} \geqslant \max \left(a_{[n, n+q']}\right)-p'$. Similarly we get $a_{k} \leqslant \min \left(a_{[n, n+q']}\right)+p'$.

Therefore, for all $k \in [n+q'-p', n+p']$ we have $a_{k} \in [\max \left(a_{[n, n+q']}\right)-p', \min \left(a_{[n, n+q']}\right)+p']$. Thus, the $(2p'-q')$-width of $n+q'-p'$ is at most $(\min \left(a_{[n, n+q']}\right)+p')-(\max \left(a_{[n, n+q']}\right)-p')=2p'-q'$. The lemma follows.

Lemma 2. Let $p'$ be a good number and $q'$ a very good number with $p'<q'$. For sufficiently large $n$, take $s, t \in [n, n+q']$ such that $\min \left(a_{[n, n+q']}\right)=a_{s}$ and $\max \left(a_{[n, n+q']}\right)=a_{t}$. Then $s \in [n, n+p']$ and $t \in [n+q'-p', n+q']$.

Proof. Lemma 2 and Lemma 3 from Solution 1 hold with $p$ and $q$ replaced by $p'$ and $q'$ by similar arguments. We can deduce the statement about $s$ from Lemma 2 of Solution 1. We can deduce the statement about $t$ from Lemma 3 of Solution 1.

Lemma 3. If $p'$ is good and $q'$ is very good with $2p'<q'$, then there exists a positive integer $r$ such that for all sufficiently large $n$, we have $a_{n+r}-a_{n} \geqslant r$.

Proof. Let $r=q'-2p'$, and let $s$ and $t$ be as defined in Lemma 2. Then consider the identity
$$\left(a_{t}-a_{n+q'-p'}\right)+\left(a_{n+p'+r}-a_{n+p'}\right)+\left(a_{n+p'}-a_{s}\right)=a_{t}-a_{s}=q'$$
By Lemma 2, we have $s \in [n, n+p']$ and $t \in [n+q'-p', n+q']$, so $a_{n+p'}-s \leqslant p'$ and $a_{t}-a_{n+q'-p'} \leqslant p'$. Combining these, we get $a_{n+p'+r}-a_{n+p'} \geqslant q'-2p'=r$. This proves that $a_{n+r}-a_{n} \geqslant r$ for sufficiently large $n$.

Lemma 4. Suppose $(p, q) \neq (1,2)$. Then there exists a good number $p'$ such that $2p'<q$.

Proof. Let $p'$ be the smallest good positive integer. Note that $p$ is good, so $p'$ exists and is less than $q$.

Suppose for contradiction that $2p' \geqslant q$. If $2p'>q$, then by Lemma 1, $2p'-q$ is a good number strictly less than $p'$, which contradicts minimality of $p'$. If $2p'=q$, then $p'<p<2p'$. So we can apply Lemma 1 with $q_{0}=p$ to get that $2p'-p$ is a good number that is strictly less than $p'$, which again contradicts minimality.

If $(p, q)=(1,2)$ then the problem is easily solved. Otherwise, Lemmas 3 and 4 combined give us some $r>0$ such that $a_{n+r}-a_{n} \geqslant r$ for $n$ sufficiently large.

By iterating, we get $a_{n+pr}-a_{n} \geqslant pr$ for all sufficiently large $n$, and hence it follows that $a_{n+p}-a_{n}=p$. Similarly we get $a_{n+q}-a_{n}=q$. As $p$ and $q$ are coprime, we deduce that $a_{n+1}-a_{n}=1$ for sufficiently large $n$. Thus we get $a_{n}=n+C$ for sufficiently large $n$, and we can conclude by the same argument as Solution 1.