Is there a simple proof for ${\small 2}\frac{n}{3}$ is not an integer when $\frac{n}{3}$ is not an integer?

Simply notice that $\rm\displaystyle\ \frac{n}3\ +\ \frac{2\:n}3\ =\ n\in \mathbb Z\ \ $ therefore $\rm\displaystyle\ \frac{n}3\in\mathbb Z\ \iff\ \frac{2\:n}3\in \mathbb Z$

This is true precisely because $\rm\:\mathbb Z\:$ is an additive subgroup of $\rm\:\mathbb Q\:,\:$ i.e. a subset closed under subtraction. For if $\rm\:S\:$ is a subgroup of a group and $\rm\ a+b\ = s \in S\ $ then $\rm\ a = b-s \in S\iff\ a+s = b\in S\:,\ $ so your property holds. Conversely if your property holds and $\rm\:a,b\in S\ $ then since $\rm\ (a-b)+b = a \in S\ $ the property implies that $\rm\: a-b\in S\:,\: $ so $\rm\:S\:$ is closed under subtraction, so $\rm\:S\:$ is a subgroup (or empty).

See also this complementary form of the subgroup property from my prior post.

THEOREM $\ $ A nonempty subset $\rm\:S\:$ of abelian group $\rm\:G\:$ comprises a subgroup $\rm\iff\ S\ + \ \bar S\ =\ \bar S\ $ where $\rm\: \bar S\:$ is the complement of $\rm\:S\:$ in $\rm\:G$

Instances of this are ubiquitous in concrete number systems, e.g.

                                 transcendental 
     algebraic * nonalgebraic  =  nonalgebraic  if  nonzero 
      rational * irrrational   =   irrational   if  nonzero 
          real *   nonreal     =    nonreal     if  nonzero 

         even  +     odd       =      odd          additive example
       integer + noninteger    =   noninteger

if $2n/3$ were an integer, then $n/3 = 4n/3 - n = 2*(2n/3) - n$ would also be an integer.

Contradiction.

Assume by contradiction that 2n/3 is an integer m and deduce from here that 3 divides n.

Hint: think about the prime factorisation of n.

Every integer $n$ can be written in the form $n = 3a + b$ where $a$ is an integer and $b$ is one of $0,1,2$. Now $n/3$ is an integer exactly when $b = 0$. Consider the representation of $2n$. There are three cases.

If $b = 0$ then $2n = 3(2a)$.

If $b = 1$ then $2n = 3(2a) + 2$.

If $b = 2$ then $2n = 3(2a+1) + 1$ since $4 = 3 + 1$.

So $2n/3$ is an integer only if $b = 0$, i.e. only if $n/3$ is an integer. And vice versa.

If (n/3) is not an integer, then n is not divisible by 3. If n is not divisible by 3, then (2*n) is not divisible by 3. Therefore, if (n/3) is not an integer, neither is ((2*n)/3).