Journey with integer steps I remember today a problem that I first hear probably 12 years ago:

Suppose that a walker starts a journey at the point $A$ in
  $\mathbb{R}^2$. The first day, the walker travels a distance of $M$ km ($M\in\mathbb{N}\setminus\{0\}$)
  and stops for the day. In the next days, the walker travels each day 1 km
  more that the previous day, in a different direction that the
  previous day, and then stops.
Question: Will the walker ever come back to the starting point? Namely, is there a path with the above conditions that ends in $(0,0)$ after an integer number $n$ of days?

For instance, I know that this problem could have solutions if the distance covered the first day is 3 km (using a Pythagorean triple $(3,4,5)$), but I don't know how to get other solutions, or even if consider $M\in\mathbb{R}$ will also be interesting.
I also don't know if this is an easy or difficult question, or if it is related with a much more difficult problem. Does anyone know how to solve this problem? or a similar one?

Post-edition: Thanks to @ChristianBlatter we know now that the answer is positive if you admit any point as a stop point. But what happen if you require furthermore that the stop points must have integer coordinates?
 A: For example, when $n=8$ the walker could go ($M$ is a positive integer):
$M$ East
$M+1$ North
$M+2$ West
$M+3$ South
$M+4$ West
$M+5$ South
$M+6$ East
$M+7$ North,
and that would bring the walker back home, while stopping only at intermediate points that have both coordinates integers.
A: This is not a full answer, but I don't have the reputation to comment yet. This problem immediately seems to be the study of intersections of circles in the plane. Whilst I can't answer in general, you can derive the (3,4,5) pythagorean triple answer using a method. 
\Say you want to return in 3 day, starting with a step of 3. On the first day you would end up somewhere on a circle, centred at the origin of radius 3, and on the start of the last day you require that you are on a circle of radius 5, origin centred. Without loss of generality you travel along the x-axis on day 1, and then draw a circle of radius 4 around (3,0). 
\The intersections of these three circles produces your known solution and proves it is the only solution on the criteria (3 days travel, starting M=3)
A: We prove the existence of such  closed walks inductively as follows:
With $(M,M+1,\ldots, M+n-1)$ we denote some admissible closed $n$-gon whose consecutive side lengths begin with $M\in{\mathbb N}_{\geq1}$.
You cannot produce an $(1,2,3)$, but you can produce an $(1,2,3,4)$ and $(M,M+1,M+2)$ for any $M\geq2$. Assume that you can do $(M,M+1,\ldots, M+n-1)$ for some  $n\geq3$. You then can replace the last leg of length $M+n-1$ by two legs of length $M+n-1$ and $M+n$ respectively, and in this way obtain an $(M,\ldots,M+n)$.
A: If I'm not wrong the random walk with vertices:
-6,  -2,
-3,  -6,   len=5
-9,  -6,   len=6
-9, -13,   len=7
-1, -13,   len=8
-1, -22,   len=9
 5, -14,   len=10
-6, -14,   len=11
-6,  -2,   len=12

fits the conditions of your edited question.
I don't have existence conditions, but, at least, we know the problem has solutions.
EDIT:
Maybe you'll like this journey, with lenghts from $1$ to $15$.

A: Here's an intuitive "proof" that such a walk exists for any $n\ge 3$. Consider the sequence $k+1$, $k+2, \dotsc, k+n$. If we regard these as the side lengths of a convex $n$-gon, then as $k\to\infty$, this approaches a regular $n$-gon, so it is certainly plausible that such an $n$-gon exists, at least for some $k$ (that is, that it can be closed).
