, 2011

Throughout this solution, the set of positive integers will be denoted by $\mathbb{Z}_{+}$.
Put $k=2+3+\cdots+2011=2023065$. We have
$$1^{n}+2 k^{n}+\cdots 2011 k^{n}=1+k \cdot k^{n}=k^{n+1}+1$$
for all $n$, so $(1, k, \ldots, k)$ is a valid sequence. We shall prove that it is the only one.
Let a valid sequence $\left(x_{1}, \ldots, x_{2011}\right)$ be given. For each $n \in \mathbb{Z}_{+}$ we have some $y_{n} \in \mathbb{Z}_{+}$ with
$$x_{1}^{n}+2 x_{2}^{n}+\cdots+2011 x_{2011}^{n}=y_{n}^{n+1}+1 .$$
Note that $x_{1}^{n}+2 x_{2}^{n}+\cdots+2011 x_{2011}^{n}<\left(x_{1}+2 x_{2}+\cdots+2011 x_{2011}\right)^{n+1}$, which implies that the sequence $\left(y_{n}\right)$ is bounded. In particular, there is some $y \in \mathbb{Z}_{+}$ with $y_{n}=y$ for infinitely many $n$.
Let $m$ be the maximum of all the $x_{i}$. Grouping terms with equal $x_{i}$ together, the sum $x_{1}^{n}+ 2 x_{2}^{n}+\cdots+2011 x_{2011}^{n}$ can be written as
$$x_{1}^{n}+2 x_{2}^{n}+\cdots+x_{2011}^{n}=a_{m} m^{n}+a_{m-1}(m-1)^{n}+\cdots+a_{1}$$
with $a_{i} \geq 0$ for all $i$ and $a_{1}+\cdots+a_{m}=1+2+\cdots+2011$. So there exist arbitrarily large values of $n$, for which
$$\begin{equation*} a_{m} m^{n}+\cdots+a_{1}-1-y \cdot y^{n}=0 . \tag{1} \end{equation*}$$
The following lemma will help us to determine the $a_{i}$ and $y$ :
Lemma. Let integers $b_{1}, \ldots, b_{N}$ be given and assume that there are arbitrarily large positive integers $n$ with $b_{1}+b_{2} 2^{n}+\cdots+b_{N} N^{n}=0$. Then $b_{i}=0$ for all $i$.
Proof. Suppose that not all $b_{i}$ are zero. We may assume without loss of generality that $b_{N} \neq 0$.
Dividing through by $N^{n}$ gives
$$\left|b_{N}\right|=\left|b_{N-1}\left(\frac{N-1}{N}\right)^{n}+\cdots+b_{1}\left(\frac{1}{N}\right)^{n}\right| \leq\left(\left|b_{N-1}\right|+\cdots+\left|b_{1}\right|\right)\left(\frac{N-1}{N}\right)^{n} .$$
The expression $\left(\frac{N-1}{N}\right)^{n}$ can be made arbitrarily small for $n$ large enough, contradicting the assumption that $b_{N}$ be non-zero. $\square$
We obviously have $y>1$. Applying the lemma to (1) we see that $a_{m}=y=m, a_{1}=1$, and all the other $a_{i}$ are zero. This implies $\left(x_{1}, \ldots, x_{2011}\right)=(1, m, \ldots, m)$. But we also have $1+m=a_{1}+\cdots+a_{m}=1+\cdots+2011=1+k$ so $m=k$, which is what we wanted to show.