For fixed $n \ge 2$ , find the maximum of real number $t$ such that $$ \sum_{i=1}^n a_i \sum_{i=1}^n \frac{1}{a_i} \ge n^2 + t \cdot \frac{\displaystyle\sum_{1\le i<j \le n} (a_i - a_j)^2 }{\left( \displaystyle\sum_{i=1}^n a_i\right)^2} $$ holds for arbitrary positive numbers $a_i\quad(i=1,2,\dots,n)$ .
This problem comes from a training math camp for CMO competitors in 2021. The standard solution seems to be a little unkind to high school students as Lagrange multiplier
method is somehow involved. I wonder if here is any solution without it. Following is my attempt.
By Lagrange identity, it's clear that
$$
\sum_{i=1}^n a_i \sum_{i=1}^n \frac{1}{a_i}=n^2 + \sum_{1\le i<j \le n}\left(\sqrt{\frac{a_i}{a_j}}-\sqrt{\frac{a_j}{a_i}}\right)^2=n^2 + \sum_{1\le i<j \le n} \frac{(a_i - a_j)^2}{a_ia_j}
$$
Then we only need to find the maximum of $t$ such that
$$
\sum_{1\le i<j \le n} \frac{(a_i - a_j)^2}{a_ia_j} \ge t \cdot \frac{\displaystyle\sum_{1\le i<j \le n} (a_i - a_j)^2 }{\left( \displaystyle\sum_{i=1}^n a_i\right)^2}
$$
However, by Cauchy-Schwarz we have
$$
\sum_{1\le i<j \le n} \frac{(a_i - a_j)^2}{a_ia_j} \ge \frac{\left(\displaystyle\sum_{1\le i<j \le n} |a_i - a_j|\right)^2}{\displaystyle\sum_{1\le i<j \le n} a_ia_j}
$$
which means nothing.
This can be treated with discussing $G = \sum_{i=1}^n \frac{1}{a_i} $. For a preparation, let's discuss cases with restrictions, which are needed later.
Define a mean $m$, where $n m = \sum_{i=1}^n{a_i}$. Then, what is a tight minimum of $G$ with fixed $m$? This can be answered immediately, since the following tight minimum is known: $n m \sum_{i=1}^n \frac{1}{a_i} = \sum_{i=1}^n a_i \sum_{i=1}^n \frac{1}{a_i} \ge n^2$, hence $G \ge \frac{n}{m}$ where the minimum is reached for all $a_i = m$.
This easy case can be expanded with another condition: Define deviations from the mean, $d_i = a_i - m$ and a variance sum $Q^2$, where $Q^2 = \sum_{i=1}^n{(a_i-m)^2}$. Then, what is a tight minimum of $G$ with fixed $m$ and $Q^2$? Obviously, the above bound does not hold, since due to the variance it is not possible that all $a_i = m$. Instead, at a tight minimum, due to the steeper rise of $\frac{1}{m + d_i}$ for negative $d_i$, one keeps those values as small as possible. So take $n-1$ values $d_i = -q$ and $d_n = (n-1) q$ to compensate for the mean condition. The variance sum condition then gives $Q^2 = \sum_{i=1}^n d_i^2 = q^2 ((n-1)^2 + n-1) = q^2 n (n-1)$. Hence,
$$G \ge \frac{n-1}{m - Q/\sqrt{n(n-1)}} + \frac{1}{m+ Q (n-1)/\sqrt{n(n-1)})}$$ is a tight lower bound. A further formal proof of this second case can be found here. If the variance sum vanishes, for $Q=0$ this falls back to the first discussed case.
After these preparations, we can now answer the main question. This can be rephrased as: find a tight minimum in
$$ F = {\left( \displaystyle\sum_{i=1}^n a_i\right)^2}\frac{\sum_{i=1}^n a_i \sum_{i=1}^n \frac{1}{a_i} - n^2}{{\displaystyle\sum_{1\le i<j \le n} (a_i - a_j)^2 }} \ge t $$
Due to homogeneity, we can demand $m=1$, which gives $$ F = n^3 \frac{\sum_{i=1}^n \frac{1}{a_i} - n}{{\displaystyle\sum_{1\le i<j \le n} (a_i - a_j)^2 }}\ge t $$
Observe that $$\sum_{1\le i<j \le n} (a_i - a_j)^2 = \frac12 \sum_{i, j=1}^n (a_i^2 + a_j^2 - a_i a_j) = n \sum_{i=1}^n a_i^2 - (\sum_{i=1}^n a_i)^2 = n \sum_{i=1}^n d_i^2 = n Q^2 \; ,$$ so we have $$ F = n^2 \frac{\sum_{i=1}^n \frac{1}{a_i} - n}{Q^2}\ge t $$ For fixed $Q^2$, after our preparatory case 2, and writing for convenience $Q^2 = n(n-1) R^2$, this is $$ F \ge F_0(R) = \frac{n}{n-1} \frac{\frac{n-1}{1 - R } + \frac{1}{1+ (n-1)R} - n}{R^2}\ge t $$ Now the minimum of $F_0$ w.r.t. $R$ occurs after a simple calculus at $R_0 = \frac{n-2}{2(n-1)}$, and upon inserting $R_0$ we obtain $$ F \ge F_0(R_0) = \frac{n}{n-1} \frac{4(n-1)^2}{n} = 4(n-1) = t $$ which is indeed the conjectured solution. It is tight, since the solution of case 2 is tight. Indeed, @RiverLi has already stated in the comments that $t = 4(n-1)$ can always be constructed, where the values for $a_i$ given by him follow the pattern of case 2. $\qquad \Box$