Given a positive integer $n \geq 2$, non-negative real numbers $a_{i}(1 \leq i \leq n)$ satisfy $\sum_{i=1}^{n} a_{i}=1$, try to find the maximum value of $\sum_{1 \leq i<j \leq n}(j-$ i) $\mathrm{a}_{\mathrm{i}} \mathrm{a}_{\mathrm{j}}$.
In summary, the maximum value of $\sum_{1 \leq \mathrm{i}<\mathrm{j} \leq \mathrm{n}}(\mathrm{j}-\mathrm{i}) \mathrm{a}_{\mathrm{i}} \mathrm{a}_{\mathrm{j}}$ is $\frac{\mathrm{n}-1}{4}$.
Solution method three: First, take $a_{1}=a_{n}=\frac{1}{2}, a_{2}=a_{3}=\ldots=a_{n-1}=0$, then $\sum_{1 \leq i<j \leq n}(j-i) a_{i} a_{j}=$ $\frac{\mathrm{n}-1}{4}$.
On the other hand, according to the conditions:
$$
\begin{array}{l}
\sum_{1 \leq i<j \leq n}(j-i) a_{i} a_{j}=\sum_{1 \leq i \leq k<j \leq n} a_{i} a_{j}=\sum_{k=1}^{n-1} \sum_{k+1 \leq j \leq n}^{1 \leq i \leq k} a_{i} a_{j} \\
=\sum_{k=1}^{n-1}\left(\sum_{1 \leq i \leq k} a_{i}\right)\left(\sum_{k+1 \leq j \leq n} a_{j}\right)=\sum_{k=1}^{n-1}\left(\frac{\left(\sum_{1 \leq i \leq k} a_{i}\right)+\left(\sum_{k+1 \leq j \leq n} a_{j}\right)}{2}\right)^{2}=\frac{n-1}{4}
\end{array}
$$

In summary, the maximum value of $\sum_{1 \leq i<j \leq n}(j-i) a_{i} a_{j}$ is $\frac{n-1}{4}$.