1. Let integer $n \geqslant 2$, set
$$
A_{n}=\left\{2^{n}-2^{k} \mid k \in \mathbf{Z}, 0 \leqslant k<n\right\} \text {. }
$$

Find the largest positive integer such that it cannot be written as the sum of one or more numbers (repetition allowed) from the set $A_{n}$.
1. First, prove that every positive integer greater than $2^{n}(n-2)+1$ can be expressed as the sum of elements in the set $A_{n}$.
Use mathematical induction on $n$.
When $n=2$, the set $A_{2}=\{2,3\}$.
For every positive integer $m$ greater than 1, we have
$$
m=\left\{\begin{array}{ll}
2+2+\cdots+2, & m \text { is even; } \\
3+\frac{m+2+\cdots+2}{\frac{m-3}{2} \uparrow}, & m \text { is odd. }
\end{array}\right.
$$

Assume the statement holds for all positive integers less than $n$. For a positive integer $n>2$, consider a positive integer $m$, $m>2^{n}(n-2)+1$.
If $m$ is even, then
$$
\begin{array}{l}
\frac{m}{2} \geqslant \frac{2^{n}(n-2)+2}{2}=2^{n-1}(n-2)+1 \\
>2^{n-1}(n-3)+1 .
\end{array}
$$

By the induction hypothesis,
$$
\frac{m}{2}=\sum_{i=1}^{r}\left(2^{n-1}-2^{k_{i}}\right),
$$

where $k_{i} \in \mathbf{Z}$, and $0 \leqslant k_{i}<n-1, i=1,2, \cdots, r$.
Thus, $m=\sum_{i=1}^{r}\left(2^{n}-2^{k_{i}+1}\right)$, which means $m$ can be expressed as the sum of elements in the set $A_{n}$.

If $m$ is odd, then
$$
\begin{array}{l}
\frac{m-\left(2^{n}-1\right)}{2} \geqslant \frac{2^{n}(n-2)+1-\left(2^{n}-1\right)}{2} \\
=2^{n-1}(n-3)+1 .
\end{array}
$$

By the induction hypothesis,
$$
\frac{m-\left(2^{n}-1\right)}{2}=\sum_{i=1}^{\prime}\left(2^{n-1}-2^{k_{i}}\right),
$$

where $k_{i} \in \mathbf{Z}$, and $0 \leqslant k_{i}<n-1, i=1,2, \cdots, r$.
Thus, $m=\sum_{i=1}^{r}\left(2^{n}-2^{k_{i}+1}\right)+\left(2^{n}-1\right)$, which means $m$ can be expressed as the sum of elements in the set $A_{n}$.

Next, prove that $2^{n}(n-2)+1$ cannot be expressed as the sum of elements in the set $A_{n}$.

Let $N$ be the smallest positive integer such that $N \equiv 1\left(\bmod 2^{n}\right)$ and it can be expressed as the sum of elements in the set $A_{n}$, i.e.,
$$
N=\sum_{i=1}^{r}\left(2^{n}-2^{k_{i}}\right),
$$

where $k_{1}, k_{2}, \cdots, k_{r} \in \mathbf{Z}$, and $0 \leqslant k_{1}, k_{2}, \cdots, k_{r}<n$.
If there exist $i, j (i \neq j)$ such that $k_{i}=k_{j}$, consider the following cases.
$$
\begin{array}{l}
\text { When } k_{i}=k_{j}=n-1, \\
N-2\left(2^{n}-2^{n-1}\right)=N-2^{n}
\end{array}
$$

can be expressed as the sum of elements in the set $A_{n}$, which contradicts the choice of $N$;

When $k_{i}=k_{j}=k<n-1$, replace these two terms with $2^{n}-2^{k+1}$, then
$$
N-2\left(2^{n}-2^{k}\right)+\left(2^{n}-2^{k+1}\right)=N-2^{n}
$$

can be expressed as the sum of elements in the set $A_{n}$, which contradicts the choice of $N$.
If all $k_{i} (i=1,2, \cdots, r)$ are distinct, then
$$
\sum_{i=1}^{n} 2^{k} \leqslant \sum_{i=0}^{n-1} 2^{i}=2^{n}-1.
$$

On the other hand, taking the modulo $2^{n}$ of both sides of equation (1), we get
$$
\sum_{i=1}^{1} 2^{k_{i}} \equiv-N \equiv-1\left(\bmod 2^{n}\right).
$$

Thus, $\sum_{i=1}^{r} 2^{k_{i}}=2^{n}-1$.
Therefore, it must be that
$$
\{0,1, \cdots, n-1\}=\left\{k_{1}, k_{2}, \cdots, k_{r}\right\}.
$$

Thus, $r=n$, and
$$
N=2^{n} n-\sum_{i=0}^{n-1} 2^{i}=2^{n}(n-1)+1.
$$

In particular, $2^{n}(n-2)+1$ cannot be expressed as the sum of elements in the set $A_{n}$.