think about how many times you will add x to the answer, when x > 1e6.This value is not too big, so we can bruteforce it.

In this formula, $$$\left\lfloor\dfrac nd\right\rfloor$$$ will be equal for some consecutive $$$d$$$

So we can calculate the left endpoint $$$l$$$ and the right endpoint $$$r$$$, then,

How to calculate $$$l$$$ and $$$r$$$?

You can find some regular patterns. Here is the conclusion: (I'm sorry that I don't know how to prove it)

$$$1$$$ is a left endpoint and for each left endpoint $$$l$$$, the right endpoint $$$r=\left\lfloor \dfrac n{\left\lfloor \dfrac nl\right\rfloor}\right\rfloor$$$.

We can do it in $$$O(\sqrt n)$$$ time.

Here is the code:

for(long long l=1,r;l<=n;l=r+1)
{
  r=n/(n/l);
  ans+=(n/l)*(l+r)*(r-l+1)/2;
}

I hope this code is correct.