Show that $\exists$ infinitely many $n\in \mathbb{N}$ such that $f(n)=f(n-1)$ where $f(n)$ is the sum of remainders when $n$ divided from $1$ to $n$ Show that there exists infinitely many positive integers $n$ such that $f(n)=f(n-1)$ where $f(n)$ is the sum of all the remainders when $n$ is divided by each of $1,2,3,\cdots, n-1,n$. 
Here are some of my observations. If $m$ divides $k$, then it leaves a remainder of $m-1$ when it divides $k-1$. Similarly, if $j$ divides $k-1$, it leaves a remainder of $1$ when it divides $k$. (Let's consider for $1<j,m<k-1$ to ease up the process). Now, including $1$, every other divisor of $k$ is coprime to that of $k-1$. 
I cannot solve it however. I have found a relation between $k$ and $k-1$ however but that might have a calculation error, so, I am not exactly sure if that holds. Here's the relation : $$2\tau{(n)}+2\tau{(n-1)}+(n-4)=\sigma{(n)}$$ where $\tau(k)$ denotes the number of divisors which divide $k$ and $\sigma(k)$ denotes the sum of all divisors of $k$ (including $1$ and $k$)
 A: Note that if $n=2^k$, then we have the remainder when $n$ is divided by a power of $2$ is $0$, and the remainder, when divided by non-powers of two, is never $0$.
If $n=2^k-1$, then we have the remainder when $n$ is divided by a power of $2$ is $2^i-1$ (for power $2^i$), and the remainder, when divided by non-powers of two, is just $1$ less than the remainder from the $n=2^k$ case.
Now, just notice that there are strictly $2^i-1$ terms between $2^i$ and $2^{i+1}$ exclusive, so we have the sum of the remainders decrease by $2^i-1$ from the power of $2$, and increase by $2^i-1$ terms all with remainder $1$ increase as we transition from $f(2^k-1)$ to $f(2^k)$ (This can be inducted on $i$ to be completely shown if necessary), so we have no overall change in the sum of the remainders.
Therefore, $f(2^k-1)=f(2^k)$.
A: What @saulspatz found to hold numerically for small values is actually true:
We have $f(2^m)=f(2^m-1)$ for all natural $m$. Here is why:
By definition, we have
$$
f(n)=\sum_{k=1}^n \left(n-k\lfloor\frac{n}{k}\rfloor\right)
$$
and thus (notice that the last term of the sum is equal to $0$)
$$
f(n)-f(n-1) = \sum_{k=1}^{n-1} \left(\left[n-k\lfloor\frac{n}{k}\rfloor\right]-\left[n-1-k\lfloor\frac{n-1}{k}\rfloor\right]\right) = \sum_{k=1}^{n-1}\left(1-k\left[\lfloor\frac{n}{k}\rfloor-\lfloor\frac{n-1}{k}\rfloor\right]\right).
$$
Now notice that $\lfloor\frac{n}{k}\rfloor-\lfloor\frac{n-1}{k}\rfloor=1$ if $k$ divides $n$, and $\lfloor\frac{n}{k}\rfloor-\lfloor\frac{n-1}{k}\rfloor=0$ otherwise. Therefore
$$
f(n)-f(n-1) = \sum_{\substack{1\leq k<n \\ k\ |\ n}}\left(1-k\right) = n-1-(\sigma(n)-n) = 2n-1-\sigma(n).
$$
Hence it sufficies to take $n$ such that $\sigma(n)=2n-1$, which can be seen to hold, for example, for $n=2^m$ with $m\in\mathbb{N}$ arbitrary.
EDIT: I'll give a bounty of $500$ reputation points and $50$ dollars to the person who proves the slightly altered version that there are infinitely many natural numbers $n$ for which $f(n)=f(n-1)-1$. Or maybe you can find an odd solution? Should be feasible, right? ;-)
