Consider a permutation $\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ of $\{1,2, \ldots, n\}$. If $a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \ldots, a_{1}+a_{2}+\ldots+a_{n}$ contains at least one perfect square, then this permutation is called a “quadratic permutation”. Find all positive integers $n$ such that every permutation of $\{1,2, \ldots, n\}$ is a “quadratic permutation”.
Thus, $x=2n+1=\frac{(3+2\sqrt{2})^{k}+(3-2\sqrt{2})^{k}}{2}$. Therefore, $n=\frac{(3+2\sqrt{2})^{k}+(3-2\sqrt{2})^{k}-2}{4}$, where $k=1,2, \mathrm{~L}, n$;
(1) If $a_{1}+a_{2}+\ldots+a_{n}$ is not a perfect square, the following permutation can make it non-"quadratic".

Let $a_{i}=i(1 \leq i \leq n), b_{k}=\sum_{i=1} a_{i}$. If $b_{k}$ is a perfect square, let $b_{k}=m^{2}$, i.e., $\frac{k(k+1)}{2}=m^{2}$,
then $\frac{k}{2}<m$, and $b_{k+1}=\frac{(k+2)(k+1)}{2}=m^{2}+k+1<(m+1)^{2}$, so $b_{k+1}$ is not a perfect square.
Swap $a_{k}$ and $a_{k+1}$, at this point, by the previous proof, the sum of the first $k+1$ terms is not a perfect square, and the sum of the first $k$ terms is $m^{2}+1$, which is also not a perfect square.

Thus, repeat the above process, i.e., for the sequence $b_{1}, b_{2}, \mathrm{~L}, b_{n}$, when a certain term is a perfect square, swap that term with the next adjacent term as described above, thereby obtaining a non-quadratic permutation, leading to a contradiction!
In summary, all positive integers $n=\frac{(3+2\sqrt{2})^{k}+(3-2\sqrt{2})^{k}-2}{4}$, where $k=1,2, \ldots, n$.

Solution: (1) If $a_{1}+a_{2}+\ldots+a_{n}$ is a perfect square, then $\frac{n(n+1)}{2}=\sum_{i=1}^{n} i=m^{2}$.
This equation can be transformed into the Pell equation: $x^{2}-2 y^{2}=1$. The solutions to this equation are:
$$
\left\{\begin{array}{l}
x_{k}+\sqrt{2} y_{k}=(3+2 \sqrt{2})^{k} \\
x_{k}-\sqrt{2} y_{k}=(3-2 \sqrt{2})^{k}
\end{array},(k=1,2, \mathrm{~L}, n)\right.
$$