For convenience we call $x_{2011}$ also $x_{0}$. Let $k$ be the largest of the numbers $x_{1}, \ldots, x_{2011}$, and consider an equation $x_{n-1}+x_{n}=2 k$, where $1 \leq n \leq 2011$. Hence we get $2 \max \left(x_{n-1}, x_{n}\right) \geq x_{n-1}+x_{n}=2 k$, so either $x_{n-1}$ or $x_{n}$, say $x_{n-1}$, satisfies $x_{n-1} \geq k$. Since also $x_{n-1} \leq k$, we then have $x_{n-1}=k$, and then also $x_{n}=2 k-x_{n-1}=2 k-k=k$. That is, in such an equation both variables on the left equal $k$. Now let $\mathcal{E}$ be the set of such equations, and let $\mathcal{S}$ be the set of subscripts on the left of these equations. From $x_{n}=k \forall n \in \mathcal{S}$ we get $|\mathcal{S}| \leq|\mathcal{E}|$. On the other hand, since the total number of appearances of these subscripts is $2|\mathcal{E}|$ and each subscript appears on the left in no more than two equations, we have $2|\mathcal{E}| \leq 2|\mathcal{S}|$. Thus $2|\mathcal{E}|=2|\mathcal{S}|$, so for each $n \in \mathcal{S}$ the set $\mathcal{E}$ contains both equations with the subscript $n$ on the left. Now assume $1 \in \mathcal{S}$ without loss of generality. Then the equation $x_{1}+x_{2}=2 k$ belongs to $\mathcal{E}$, so $2 \in \mathcal{S}$. Continuing in this way we find that all subscripts belong to $\mathcal{S}$, so $x_{1}=x_{2}=\cdots=x_{2011}=k$.