Is every odd integer greater than $105$ the sum of two coprime composite numbers? 
Can any odd integer greater than 105 be represented as the sum of two
   coprime composite numbers?

The answer to this question conjectures that any integer greater than $210$ is the sum of two coprime composite numbers. Numbers as sum of two relatively prime composite numbers
 A: Claim: For an odd integer $n\geq 107$, there exists an odd prime $p$ such that $p \not\mid n$ and $p^2+4\leq n$.
Proof: Assume the contrary. Then for all odd primes under $\sqrt{n-4}$, $p \mid n$. $$\prod_{p<\sqrt{n-4}}p | n$$
Now by Bertrand's postulate, there exists a prime in each of the intervals $[8,16),[16,32),\cdots, [2^k,2^{k+1})$, and also $3,5,7$.
We have $2^{k+1}\leq\sqrt{n-4} \implies k = \frac{1}{2}\log_2(n-4)-1$
For $n\geq 260$, we have $k \geq 3$
A lower bound we can achieve on the product of them will be
$$\prod_{p < \sqrt{n-4}}p \geq 3\cdot5\cdot7\cdot2^{3+4+\cdots+k}=\frac{105}{8}\cdot 2^{\frac{k(k+1)}{2}}\geq^* \frac{105}{8}\cdot 2^{2k-\frac{1}{2}}=\frac{105}{8}\cdot 2^{\log_2(n-4)-2}=\frac{105(n-4)}{8}>n$$
And thus the produtc doesn't divide $n$, contraditcion.
For values $107 \leq n \leq 259$, $\sqrt{n-4}\geq 10$, and the product of odd primes under $n$ are $3\cdot5\cdot7=105$. Clearly one of $3,5,7$ doesn't divide $n$ unless $n=210$, for which $11$ doesn't divide $n$.

Then $p^2$ and $n-p^2$ are composite numbers and they're coprime. ($n-p^2$ is even and $\geq 4$)
Edit: Actually after rereading my "thing", I don't see any reason why this can't be applied to a even number - just replace $4$ with $9$.
