2. Denote by $\left(x_{k}, y_{k}\right)$ the pair obtained after $k$ steps. The sum $x_{k}+y_{k}$ is invariant and equals $s=n+b$. Since $2 \cdot\left(x+\frac{y}{2}\right) \equiv 2 \cdot \frac{x}{2} \equiv x(\bmod x+y)$, we have $2 x_{k} \equiv x_{k-1}$ $(\bmod s)$. A simple induction yields

$$2^{k} x_{k} \equiv x_{0}=n \quad(\bmod s)$$

Since $\left(s, 2^{k}\right)=1$, it is enough to prove the existence of an odd number $s, n<s<$ $2 n$, such that for some $k$ we have $2^{k} b \equiv n(\bmod s)$, i.e. $\left(2^{k}+1\right) n \equiv 0(\bmod s)$. To this end, one can simply take $s=2^{r}+1$ and $k=r$, where $2^{r-1}<n<2^{r}(r \in \mathbb{N})$. Thus $b=2^{r}+1-n$.

Remark. Clearly, one can take any $s$ such that $s \mid 2^{k}+1$ for some $k \in \mathbb{N}$. For example, $s=3^{i} 11^{j}\left(i, j \in \mathbb{N}_{0}\right)$ works. From here, one can deduce that, given any constants $0<\alpha<\beta$, for all big enough $n$, there is a desired number $b$ with $\alpha n<b<\beta n$.