The sum of consecutive integers is $50$. How many integers are there? I started off by calling the number of numbers in my list "$n$". Since the integers are consecutive, I had $x + (x+1) + (x+2)...$ and so on. And since there were "$n$" numbers in my list, the last integer had to be $(x+n)$. This is where I got stuck. I didn't know how to proceed because I am not given the starting point of my integers, nor an ending point.
If your $n$ is odd, then the middle number has to be $50/n$.  The odd divisors of $50$ are $1$ and $5$, which gives us two solutions $50=50$ and $8+9+10+11+12=50$

If $n$ is even, then $50/n$ is the half-integer between the middle two numbers.  So $n$ has to be an even divisor of $100$, but not a divisor of $50$, so $n=4$ or $20$.  

If $n=4$, then $50/4 = 12.5$ and we get $11+12+13+14=50.$

If $n=20$. then $50/20 = 2.5$ and we get $-7+-6+-5+\cdots +11+12 = 50.$

So there are 4 answers: $n=1, 4, 5, $ and $20$.

Edit:  As Bill points out, I missed the divisors $25$ and $100$, which give two more answers:  $50 = -10+-11+\cdots+14$ and $50 = -49 +-48+\cdots +50$.

Note that each solution with negative integers is related to an all-positive solution.  From the solution $11+12+13+14=50$, we just prepend the terms $-10, -9, \ldots, 10$, which add to $0$, and we have another solution.
First, welcome to Mathematics stack exchange. Second, if you start like $x+(x+1)+...+(x+n)$, you will end nowhere. $50$ is a concrete and small number so what can you try is to use that fact, sum of two consecutive numbers it is not, due to sum of two consecutive numbers can not be divided with $2$ in $\mathbb{Z}$, sum of three consecutive numbers is divisible with $3$, so it is not sum of three numbers. Four consecutive numbers- now this is on first look possible(remainder mod $4$ is $2$) but each one of them has to be around $12$, so first options that you have is $11+12+13+14=23+27= 50$, hence the solution. Maybe there are even other solution, but I feel free to interpret your question as required to see one possible solution.