11. (20 points) Find all positive integers $n \geqslant 3$, such that the following proposition holds: If $a_{1}, a_{2}, \cdots, a_{n}$ form an arithmetic sequence, and if $a_{1}+2 a_{2}+\cdots+n a_{n}$ is a rational number, then at least one of $a_{1}, a_{2}, \cdots, a_{n}$ is a rational number.
11. Let the common difference of the sequence $a_{1}, a_{2}, \cdots, a_{n}$ be $d$, then
$$
\begin{array}{l} 
a_{1}+2 a_{2}+\cdots+n a_{n}= \sum_{k=1}^{n} k\left(a_{1}+(k-1) d\right)=a_{1} \sum_{k=1}^{n} k+d \sum_{k=1}^{n} k(k-1) \\
= \frac{n(n+1)}{2} a_{1}+\frac{d}{3} \sum_{k=2}^{n}((k+1) k(k-1)-k(k-1)(k-2)) \\
= \frac{n(n+1)}{2} a_{1}+\frac{(n+1) n(n-1)}{3} d \\
=\frac{n(n+1)}{2}\left(a_{1}+\frac{2(n-1)}{3} d\right) .
\end{array}
$$

If $3 \mid n-1$, then from the above formula, we have $a_{1}+2 a_{2}+\cdots+n a_{n}=\frac{n(n+1)}{2} a_{\frac{2 n+1}{3}}$. Therefore, if $a_{1}+2 a_{1}+\cdots+n a_{n}$ is a rational number, then $a_{\frac{2 n+1}{3}}$ is a rational number. On the other hand, if $3 \nmid n-1$, let $a_{1}=\sqrt{2}, d=-\frac{3}{2(n-1)} \sqrt{2}$, then each term in the arithmetic sequence is irrational, while $a_{1}+2 a_{2}+\cdots+n a_{n}=0$ is a rational number.
In summary, the set of positive integers $n$ that satisfy the condition is $\left\{n \in \mathbf{N}^{*} \mid n \equiv 1(\bmod 3)\right\}$.