Proof read from "A problem seminar" May you help me judging the correctness of my proof?:
Show that the if $a$ and $b$ are positive integers, then
$$\left(a+\frac{1}{2}\right)^n+\left(b+\frac{1}{2}\right)^n$$
is integer for only finintely many positive integers $n$
We want $n$ so that
$$\left(a+\frac{1}{2}\right)^n+\left(b+\frac{1}{2}\right)^n\equiv0\pmod{1}$$
So we know by the binomial theorem that
$(an+b)^k\equiv b^n\pmod{n}$ for positive $k$
Then,
$$\left(a+\frac{1}{2}\right)^n\equiv(1/2)^n\pmod{1}$$
and similarly with the $b$
So 
$$\left(a+\frac{1}{2}\right)^n+\left(b+\frac{1}{2}\right)^n\equiv 2*(1/2)^n\pmod{1}$$
Therefore, we want $2*(1/2)^n$ to be integer, so that $2^n|2$
clearly, the only positive option is $n=1$
(Editing, my question got prematurely posted. Done)
 A: Our expression can be written as 
$$\frac{(2a+1)^n+(2b+1)^n}{2^n}.$$
If $n$ is even, then $(2a+1)^n$ and $(2b+1)^n$ are both the squares of odd numbers. 
Any odd perfect square is congruent to $1$ modulo $8$. So their sum is congruent to $2$ modulo $8$, and therefore cannot be divisible by any $2^n$ with $n\gt 1$.
So we can assume that $n$ is odd. For odd $n$, we have the identity 
$$x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+\cdots +y^{n-1}).$$
Let $x=2a+1$ and $y=2b+1$. Note that $x^{n-1}-x^{n-2}y+\cdots +y^{n-1}$ is a sum of an odd number of terms, each odd, so it is odd.
Thus the highest power of $2$ that divides $(2a+1)^n+(2b+1)^n$ is the highest power of $2$ that divides $(2a+1)+(2b+1)$. Since $(2a+1)+(2b+1)\ne 0$, there is a largest $n$ such that our expression is an integer.
Remark: The largest $n$ such that our expression is an integer can be made quite  large.    You might want to see for example what happens if we  let $2a+1=2049$ and $2b+1=2047$. Your proposed proof suggests, in particular, that $n$ cannot be greater than $1$.  
I suggest that when you are trying to write out a number-theoretic argument, you avoid fractions as much as possible and deal with integers only. 
A: Andre gave you a solution, but you wanted proofreading, so I'll point out two mistakes you've made (I'm not saying there aren't more :-)).
First, modulo arithmetic assumes you're working with integers, so it cannot be applied on rationals.
Second, when getting from
$$(an+b)^n\equiv b^n\pmod{n}$$
to
$$\left(a+\frac{1}{2}\right)^n\equiv(1/2)^n\pmod{1}$$
You put $n = 1$, but just in some spots (but left it unchanged in the exponents). You cannot use formulas like that. It's like going from
$$n = n$$
and then substituting the first $n$ with $1$, getting
$$n = 1.$$
Of course you'll get $n = 1$ when you use it like that.
