How do I solve Problem 1 of the International Olympiad of Metropolis? The question is: Find all $n$ so that there exists $n$ consecutive numbers whose sum is a square!

My method to solve the problem: I would try to look at the values modulo $n$, and there I see there is no possible square for $n=2x&gt;2$, so all even numbers bigger than 2, and that all odd numbers work...

I can also see that the sum of the consecutive numbers is $$\dfrac{((x+1)+(x+n))
\cdot n}{2}$$ So that has to give a square...

But how do I do that? I cannot put my method in application... Am I totally wrong, and I shouldn't do it like that?
As you noted in the question, if $x +1$ is the first number in the sum, then the sum of the $n$ consecutive numbers is
$$ \frac{n}{2} \cdot (x + 1 + x + n). $$

If $n$ is odd, let $n = 2k + 1$, and take $x = k$. Then the sum of the $n$ consecutive numbers $(k + 1), (k + 2), \dots, (k + n)$ is
$$
  \frac{2k+1}{2} \cdot (k + 1 + k + 2k + 1) = (2k + 1)^2
$$
which is a square, so all odd numbers work.

Now suppose that $n$ is even. Let $n = 2k$ for some $k$.

Then the sum is equal to
$$ k(2x + 2k + 1) $$.

The term in brackets is odd, so for this to be a square, the largest power of $2$ which divides $k$ must be even. i.e. We must have that $k = 2^{2m} \cdot d$ where $m$ is some natural number, and $d$ is an odd natural number.

The sum then reduces to
$$ 2^{2m} \cdot d \cdot (2x + 2^{2m + 1} \cdot d + 1) $$

This is a square if and only if
$$ d \cdot (2x + 2^{2m + 1} \cdot d + 1) $$
is a square, so it is enough to choose $x$ so that the term in brackets is a square multiple of $d$. That is, we try to choose $x$ so that
$$ 2x + 2^{2m + 1} \cdot d + 1 = a^2 \cdot d $$
for some natural number $a$.

This is equivalent to
$$ 2x + 1 = \left( a^2 - 2^{2m + 1} \right) \cdot d $$
and it is easy to see that this has a solution for x as long as $a$ is an odd square greater than $2^{2m + 1}$. Thus we have a solution in this case since there are arbitrarily large odd square numbers.

We see that if $n$ is even, then $n$ works if and only if $n$ is divisible by $2$ an odd number of times.
For any odd $n$, sum of $n$ consecutive numbers starting from $\frac {n+1}2$ is 

$$\frac n2 (n+1+(n-1)*1) = n^2$$

Thus for every odd $n$, there exist $n$ consecutive numbers whose sum is a perfect square.

More details - This becomes simpler if you know Arithematic Progression. The formula for sum of $n$ numbers starting from $a$, which are $d$ apart (difference between successive numbers is $d$) is 

$$S_n=\frac n2(2a+(n-1)d)$$

Here $d=1$ and $a$ can be calculated to be $\frac {n+1}2$

Thank you @Litho for pointing it out.