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 $\sigma(N+1)$.