IMO Selection

Solution:

Working in any base $d \geq 2$, by pigeonhole, two of the numbers
$$1, 11, 111, \ldots, \overbrace{11 \cdots 11}^{n+1}$$
have the same residue $\bmod n$. Therefore their difference
$$D = \overbrace{11 \cdots 11}^{t} \overbrace{00 \cdots 00}^{k} \text{ with } k, t \leq n$$
is divisible by $n$ in base $d$. But then also
$$n \mid D \cdot d^{n-k} \cdot \left(1 + d^{t} + d^{2t} + \cdots + d^{n!-t}\right) = \overbrace{11 \cdots 11}^{n!} \overbrace{00 \cdots 00}^{n}$$
in base $d$. But the representation of this number is independent of $d$, hence we have found our sequence of zeros and ones.

Solution:

It suffices to find a non-empty, finite set $\sigma \in \mathbb{N}$ such that
$$n \mid \sum_{s \in \sigma} d^{s}$$
for all $d \geq 2$. Consider the set $\sigma = \{\varphi(n), 2\varphi(n), \ldots, n\varphi(n)\}$. For $d \geq 2$, if $(d, n) = 1$:
$$\sum_{s \in \sigma} d^{s} = \sum_{k=1}^{n} \left(d^{\varphi(n)}\right)^{k} = n$$
Else, if $g = (d, n) \neq 1$, let $n = x y$ with $(x, y) = (y, d) = 1$ and $y$ maximal:
$$\sum_{s \in \sigma} d^{s} = \sum_{k=1}^{n} \left(d^{\varphi(n)}\right)^{k} \equiv \frac{d^{\varphi(n)}\left(d^{n\varphi(n)}-1\right)}{d^{\varphi(n)}-1} \equiv 0 \quad \bmod x$$
which makes sense, as $d^{\varphi(n)}-1$ is coprime to $x$, and true since $v_{p}(x) \leq v_{p}(n) \leq \varphi(n)$ (ask raphi for an elaborate explanation). And $y$ divides this sum by the first argument as above. Now $n = x y$ divides the sum since $(x, y) = 1$.