Selection tests for the International Mathematical Olympiad, 2013

We will construct such a sequence by induction:
Define $a_{1}=1$ and $a_{2}=2$. In this case we have $b_{1}=|a_{1}-a_{2}|=1$.
Assume that $a_{1}, a_{2}, \ldots, a_{2n}$ are defined such that there is no positive integer which occurs at least twice neither in the finite sequence $a_{1}, a_{2}, \ldots, a_{2n}$ nor in the finite sequence $b_{1}=|a_{1}-a_{2}|, b_{2}=|a_{2}-a_{3}|, \ldots, b_{2n-1}=|a_{2n-1}-a_{2n}|$.
Let $M_{n}$ be the maximum element of the set of integers $\{a_{1}, a_{2}, \ldots, a_{2n}\}$, $c_{n}$ the maximum of the integers $k$ such that $\{1,2, \ldots, k\} \subseteq \{a_{1}, a_{2}, \ldots, a_{2n}\}$, and $d_{n}$ the maximum of the integers $k$ such that $\{1,2, \ldots, k\} \subseteq \{b_{1}, b_{2}, \ldots, b_{2n-1}\}$.
Notice that for all $1 \leq i \leq 2n-1$, we have
$$b_{i}=|a_{i}-a_{i+1}| \leq \max_{1 \leq j \leq 2n} a_{j} - \min_{1 \leq j \leq 2n} a_{j} = M_{n}-1.$$
If $c_{n}<d_{n}$ define $a_{2n+1}=2M_{n}+c_{n}+1$ and $a_{2n+2}=c_{n}+1$. If $c_{n} \geq d_{n}$ define $a_{2n+1}=2M_{n}+1$ and $a_{2n+2}=2M_{n}+d_{n}+2$.
It is clear that none of the two possible values for $a_{2n+1}$ and for $a_{2n+2}$ occur in the finite sequence $a_{1}, a_{2}, \ldots, a_{2n}$.
In the first case, we have
$$b_{2n}=2M_{n}+c_{n}+1-a_{2n} \geq M_{n}+c_{n}+1 > M_{n}-1 \geq b_{i}$$
and
$$b_{2n+1}=2M_{n} > M_{n}-1 \geq b_{i}$$
for all $1 \leq i \leq 2n-1$, and $b_{2n+1} \neq b_{2n}$ since $a_{2n} \neq c_{n}+1$.
In the second case, we have
$$b_{2n}=2M_{n}+1-a_{2n} \geq M_{n}+1 > M_{n}-1 \geq b_{i}$$
and
$$b_{2n+1}=d_{n}+1 \neq b_{i}$$
for all $1 \leq i \leq 2n-1$, and $b_{2n+1}=d_{n}+1 \leq M_{n} < M_{n}+1 \leq b_{2n}$.
Therefore, there is no positive integer which occurs at least twice in the finite sequence $a_{1}, a_{2}, \ldots, a_{2n+2}$ or in the finite sequence $b_{1}, b_{2}, \ldots, b_{2n+1}$.
This proves that there is no positive integer which occurs at least twice in the infinite sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ or in the infinite sequence $b_{1}, b_{2}, \ldots, b_{n}, \ldots$.
Notice moreover, that in the first case $c_{n+1} \geq c_{n}+1$ and in the second case $d_{n+1} \geq d_{n}+1$. Therefore, if $e_{n}=\min\{c_{n}, d_{n}\}$ then $e_{n+2} \geq e_{n}+1$. But $c_{1}=2$ and $d_{1}=1$. We deduce that $e_{2n} \geq n$ for all integer $n$. Therefore, any positive integer $n$ occurs in both finite sequences $a_{1}, a_{2}, \ldots, a_{4n}$ and $b_{1}, b_{2}, \ldots, b_{4n-1}$.
This proves that any positive integer $n$ occurs exactly once in each infinite sequence $a_{1}, a_{2}, \ldots$ and $b_{1}, b_{2}, \ldots$.