If $N$ is a positive integer then,$$\sum\limits_{n=1}^{N}\sigma(n)=\sum\limits_{n=1}^{N}n\lfloor\dfrac{N}{n}\rfloor$$, where $\lfloor.\rfloor$ denotes greatest integer function and $\sigma(n)$ denotes sum of divisors of $n$ (positive).
I can't think for it, please give very elementary proof here.
On
Let's use induction on $N$. For $N=1$ the equality becomes $1=1$.
For $n=1,\ldots, N$, $n\lfloor N/n\rfloor$ is the greatest multiple of $n$ that is not greater than $N$.
Then, assuming that the equality holds, let's try to prove that $$\sum_{n=1}^{N+1}\sigma(n)=\sum_{n=1}^{N+1}n\left \lfloor \frac {N+1}n\right\rfloor$$
LHS has increased $\sigma(N+1)$. RHS is the sum of the greatest multiples of $n$ that are not greater than $N+1$. That is, the difference between this sum and the former is that now we are summing $N+1$ when we were summing $N+1-n$, and that happens when $n$ is a divisor of $N+1$. So the difference is also $\sigma(N+1)$.
In symbols, $$\begin{align} \sum_{n=1}^{N+1}&n\left \lfloor \frac {N+1}n\right\rfloor-\sum_{n=1}^{N}n\left \lfloor \frac {N}n\right\rfloor\\ &=\sum_{n=1}^{N+1}n\left \lfloor \frac {N+1}n\right\rfloor-\sum_{n=1}^{N+1}n\left \lfloor \frac {N}n\right\rfloor\\ &=\sum_{n=1}^{N+1}n\left(\left \lfloor \frac {N+1}n\right\rfloor-\left \lfloor \frac {N}n\right\rfloor\right)\\ \end{align}$$
The difference in parentheses is $1$ when $n\mid N+1$ and $0$ otherwise. Then the sum is $\sigma(N+1)$.
First, we have $$\sum_{n\leq N} \sigma(n)=\sum_{n\leq N}\sum_{d|n}d.$$ Now, with the change $n=dq$, the double sum becomes $$\sum_{n\leq N}\sum_{d|n}d=\sum_{dq\leq N}d =\sum_{d\leq N}\sum_{q\leq N/d}d =\sum_{d\leq N}d\sum_{q\leq N/d}1 =\sum_{d\leq N}d\lfloor N/d\rfloor$$ as wanted.