$f(n) + f(n+1) + ... + f(n+N-1) = n$ 
Let $N \in \mathbb{N}$ be constant. Find all non-decreasing functions $f:\mathbb{Z} \rightarrow \mathbb{Z}$ satisfying
  $$f(n) + f(n+1) + ... + f(n+N-1) = n$$
  for all integers $n$.

Please check my solution.
$$
f(n) + f(n+1) + ... + f(n+N-1) = n \tag{1}
$$ 
$$
f(n+1) + f(n+2) + ... + f(n+N) = n+1 \tag{2}
$$
(2)-(1): 
$$
f(n+N) - f(n) = 1
$$
As $f$ are non-decreasing function, so $\{f(n+1), f(n+2), ..., f(n+N-1)\} \in \{f(n)$ or $f(n)+1\}$
If $f(n+l), f(n+l+1), ...,f(n+N-1) = f(n)+1$ 
and $f(n+l)$ has the smallest value that is equal to $f(n)+1$ 
then $f(n+l-N), f(n+l+1-N), ..., f(n-1) = f(n)$ 
and $f(n), f(n+1), ..., f(n+l-1) = f(n)$ 
There would exist $k\in \mathbb{N}$ such that  $f(k), f(k+1), ..., f(k+N-1)$ are all equal
and the solution would be $k\cdot f(n)+(N-k)(f(n)+1)=n$
i.e., $f(n) = \frac{n-N+k}{N}$, Contradiction
Therefore, $\{f(n+1), f(n+2), ... , f(n+N-1)\}$ are all equal where $n \equiv 0 (\bmod N)$
My answer is $f(n) = \left\lfloor \frac{n}{N} \right\rfloor$ where $n\in \mathbb{Z}$
Check : 
Let $\left\lfloor \frac{n}{N} \right\rfloor = l, n = Nl+t$
$\Rightarrow f(n)=f(n+1)=...=f(n+N-2-t)=l$
$f(n+N-1-t)=f(n+N-1) = ...=f(n+N-1)$
so $f(n)+f(n+1)+...+f(n+N-1) = l(N-1-t) + (l+1)(1+t) = Nl +t $
 A: Your solution for $k$ is not very clear and I don't think it's entirely correct:

There would exist $k\in\mathbb N$ such that $f(k),f(k+1),...,f(k+N−1)$
  are all equal and the solution would be $k⋅f(n)+(N−k)(f(n)+1)=n$
i.e., $f(n) = \frac{n-N+k}{N}$, Contradiction

It looks like the factors $k$ and $(N-k)$ is the numder of terms that are $f(n)$ and $f(n+1)$ respectively. But that can't be like that since that number would depend on $n$. (Besides a contradiction here would indicate that we have no solutions as the existence of the solution is basically the only hypothesis we've introduced).
Let's step back and see what we have. We have concluded that there is a $k$ such that $f(k+j)$ is constant ($0\le j<N$), we can also see that $f(n+jN) = f(n)+j$. So what we basically have is a function that steps one step every $N$ integers. We also see by this that we can have $0\le k< N$ and that $f$ is a function on the form
$$f(n) = \left\lfloor {n-k\over N}\right\rfloor + c$$
Now to determine $k$ and $c$ we consider the sum
$$\sum_0^{N-1}f(j) = 0$$ 
We use that $f(j) = c-1$ if $j<k$ and $c$ otherwise so we have 
$$\sum_0^{N-1}f(j) 
= k(c-1) + (N-k)c 
= Nc - k = 0$$
Now since $0\le k<N$ we have $-Nc \le k-Nc = 0 < N-Nc$ and especially that $Nc \ge 0 > N(c-1)$ So we have $1>c\ge 0$ so $c=0$ and therefore also $k=0$.
So at least your conclusion is correct, the only such function is:
$$f(n) = \left\lfloor{n\over N}\right\rfloor$$
