IMO Shortlist

For every indices $m \leqslant n$ we will denote $S(m, n)=a_{m}+a_{m+1}+\cdots+a_{n-1}$; thus $S(n, n)=0$. Let us start with the following lemma.

Lemma. Let $b_{0}, b_{1}, \ldots$ be an infinite sequence. Assume that for every nonnegative integer $m$ there exists a nonnegative integer $n \in [m+1, m+r]$ such that $b_{m}=b_{n}$. Then for every indices $k \leqslant \ell$ there exists an index $t \in [\ell, \ell+r-1]$ such that $b_{t}=b_{k}$. Moreover, there are at most $r$ distinct numbers among the terms of $(b_{i})$.

Proof. To prove the first claim, let us notice that there exists an infinite sequence of indices $k_{1}=k, k_{2}, k_{3}, \ldots$ such that $b_{k_{1}}=b_{k_{2}}=\cdots=b_{k}$ and $k_{i}<k_{i+1} \leqslant k_{i}+r$ for all $i \geqslant 1$. This sequence is unbounded from above, thus it hits each segment of the form $[\ell, \ell+r-1]$ with $\ell \geqslant k$, as required.

To prove the second claim, assume, to the contrary, that there exist $r+1$ distinct numbers $b_{i_{1}}, \ldots, b_{i_{r+1}}$. Let us apply the first claim to $k=i_{1}, \ldots, i_{r+1}$ and $\ell=\max \{i_{1}, \ldots, i_{r+1}\}$; we obtain that for every $j \in \{1, \ldots, r+1\}$ there exists $t_{j} \in [s, s+r-1]$ such that $b_{t_{j}}=b_{i_{j}}$. Thus the segment $[s, s+r-1]$ should contain $r+1$ distinct integers, which is absurd.

Setting $s=0$ in the problem condition, we see that the sequence $(a_{i})$ satisfies the condition of the lemma, thus it attains at most $r$ distinct values. Denote by $A_{i}$ the ordered $r$-tuple $(a_{i}, \ldots, a_{i+r-1})$; then among $A_{i}$'s there are at most $r^{r}$ distinct tuples, so for every $k \geqslant 0$ two of the tuples $A_{k}, A_{k+1}, \ldots, A_{k+r^{r}}$ are identical. This means that there exists a positive integer $p \leqslant r^{r}$ such that the equality $A_{d}=A_{d+p}$ holds infinitely many times. Let $D$ be the set of indices $d$ satisfying this relation.

Now we claim that $D$ coincides with the set of all nonnegative integers. Since $D$ is unbounded, it suffices to show that $d \in D$ whenever $d+1 \in D$. For that, denote $b_{k}=S(k, p+k)$. The sequence $b_{0}, b_{1}, \ldots$ satisfies the lemma conditions, so there exists an index $t \in [d+1, d+r]$ such that $S(t, t+p)=S(d, d+p)$. This last relation rewrites as $S(d, t)=S(d+p, t+p)$. Since $A_{d+1}=A_{d+p+1}$, we have $S(d+1, t)=S(d+p+1, t+p)$, therefore we obtain
$$a_{d}=S(d, t)-S(d+1, t)=S(d+p, t+p)-S(d+p+1, t+p)=a_{d+p}$$
and thus $A_{d}=A_{d+p}$, as required.

Finally, we get $A_{d}=A_{d+p}$ for all $d$, so in particular $a_{d}=a_{d+p}$ for all $d$, QED.