Question: Notation: let us have a particle of spin(more generally angular momentum, but for simplicity just consider spin) $j\_1$ with projection to the third axis $m\_1$ and a particle with spin $j\_2$ with projection of $m\_2$. We want to combine them into a doublet with total spin $J$ and its projection $M$. We write clebsch-gordan coefficient $( j\_1 m\_1 j\_2 m\_2 | J M )$.

Now consider we have two clebsh-gordan coefficients that we want to multiply:
$$\sum\_{M\in\{-J,..,J\}}( j\_1 m\_1 j\_2 m\_2 | J M )(j'\_1 m'\_1 j'\_2 m'\_2 | J M )$$
my question is: Is there a simple formula this can be reduced into?

Answer: 
Quoting Wikipedia on Clebsch-Gordan coefficients we have the orthogonality relation:

$$\sum^{j\_1+j\_2}\_{J=|j\_1-j\_2|}\sum^{J}\_{M=-J}<j\_1,m\_1,j\_2,m\_2|J,M><J,M|j\_1,m'\_1,j\_2,m'\_2>=\\=<j\_1,m\_1,j\_2,m\_2|j\_1,m'\_1,j\_2,m'\_2>=\delta\_{m\_1,m'\_1}\delta\_{m\_2,m'\_2}$$

This is the closest I could find resembling what you presented above. I hope this helps.
The simplest way to look at such sums is to start with
$$
\langle j\_1m\_1; j\_2m\_2\vert j\_1’m\_1’; j\_2’m\_2’\rangle =
\delta\_{j\_1j\_1’}\delta\_{j\_2j\_2’}\delta\_{m\_1m\_1’}\delta\_{m\_2m\_2’}\, ,
$$
insert the unit operator in the form of
$$
1= \sum\_{JM}\vert JM\rangle\langle JM\vert
$$
where $J$ runs over all the allowed value in the coupling $j\_1\otimes j\_2$, $M$ runs from $-J$ to $J$, so as 
to get
$$
\sum\_{JM}\langle j\_1m\_1; j\_2m\_2\vert JM\rangle \langle JM\vert 
j\_1’m\_1’; j\_2’m\_2’\rangle =
\delta\_{j\_1j\_1’}\delta\_{j\_2j\_2’}\delta\_{m\_1m\_1’}\delta\_{m\_2m\_2’}
\tag{1}
$$
and recall that the CGs are real, i.e. 
$$
\langle JM\vert j\_1’m\_1’;j\_2’m\_2’\rangle = 
\langle j\_1’m\_1’;j\_2’m\_2’\vert JM\rangle \tag{2}\, .
$$
Inserting (2) in (1) gives the desired orthogonality relation.