is any faster algorithm to calculate this series? ∑∑ai​/aj
Assuming that your numbers $a_i$ are small enough and $n$ is large enough, the following algorithm may work better in practice.
For each $a_i$, for all possible division results $d_j$, find minimum $v_j$ such that $\lfloor \frac {a_i} {v_j} \rfloor = d_j$. The idea is the following:

There can't be too many different $d_j$. E.g. for $a=10^9$ there are about $6 \cdot 10^4$ possible $d_j$. It's possible to show that there are $O(\sqrt{a})$ possible values of $d_j$ (about $2 \sqrt a$). Note that it's possible to iterate all possible $d_j$'s in time linear of their amount.

All numbers between neighboring $v_j$'s give the same result after division.


Therefore, the algorithm is the following:

Sort $a_1, \ldots, a_n$
For each $a_i$, find all possible $\{(d_j, v_j)\}$ as described above. Using binary search on array of $a_i$, for all $j$ find the number of $a_i$ lying between $v_j$ and $v_{j+1}$. Division by any of them results in $d_j$.

P.S.: This algorithm can take exponential time (since it's polynomial on the numbers themselves), but it can have a guaranteed $O(n^2 \log n)$ time when we ignore $d_j$ which can't be achieved: when we've found the last number $a_k$ which lies between $v_j$ and $v_{j+1}$, we select the next $d$ to consider based on $a_{k+1}$. Additional $\log n$ comes from the binary search.