The short answer is (like Andres Caicedo pointed out) that here $v$ and $w$ are fixed, and the sum ranges over such pairs $p$ and $q$ that satisfy the condition $p+q=v+w$. But they are not necessarily integers. The weights are elements of a certain free abelian group, so that is the reason the sum is not written in the usual form, where a lower and an upper bound is given.
This is partly educated guessing, but the context you give (here and in your other question) screams for the following interpretation: $V$ and $W$ are representations of your (presumably semisimple) Lie-algebra. This sum arises in a study of their tensor product $V\otimes W$. The underlying observation is that if $x\in V$ is of weight $p$ and $y\in W$ is of weight $q$, then their tensor product $x\otimes y$ is of weight $p+q$. Therefore all of $V_p\otimes W_q$ is contained in the weight space $(V\otimes W)_{p+q}$. The dimension of the tensor product is $dim (V_p\otimes W_q) = dim V_p \,dim W_q.$
The question this formula seeks to answer, is to compute the dimension of a weight space $(V\otimes W)_\mu$. If $\mu$ is a weight of the tensor product representation, then it must be of the form $\mu=v+w$, where $v$ is a weight of $V$ and $w$ is a weight $W$. But more often than not we can write $\mu$ in several ways as a sum $v+w=\mu=p+q$ of two weights. So all such tensor products of weight spaces $V_p\otimes W_q\subset (V\otimes W)_{\mu}.$ All the representations here are direct sums of their weight spaces, so the sum of these tensor products of subspaces is also direct. Therefore $dim (V\otimes W)_{v+w}$ is the sum of the dimensions of the tensor products of the individual weight spaces $V_p$ and $W_q$ subject to the constraint $p+q=v+w$.