Yes. For example, with quiver representations, we have a formula
chi(M,N)=dim Hom(M,N)-dim Ext1(M,N) = \sum di(M)di(N) - \sumi -> j di(M)di(N).$\chi(M,N)=\dim Hom(M,N)-\dim Ext^1(M,N) = \sum d_i(M)d_i(N) - \sum_{i \to j} d_i(M)d_j(N).$
where di(M)$d_i(M)$ is the dimension of M at node i. The proof is to check that it's true for simples, and then note that the category of representations of the path algebra of a quiver has global dimension 1.
So, what you've noted above is that this is positive definite if and only if the graph is Dynkin. Well, what's good about being positive definite? For one thing, if an object has trivial Ext^1 with itself, then it is rigid, it has no deformations. On the other hand, it also must have chi(M,M)>0$\chi(M,M)>0$, since Hom always has positive dimension, and Ext1(M,M)=0$Ext_1(M,M)=0$.
Thus, if our quiver is not Dynkin, it has dimension vectors where no module can be rigid. On the other hand, if you work a bit harder, you can show that if the graph is Dynkin, a dimension vector corresponds to a unique indecomposible module if and only if chi(M,M)=1, which is exactly the positive roots of the root system. This is Gabriel's theorem.:
if the graph is Dynkin, every dimension vector has a unique rigid module and this is indecomposible if and only if $\chi(M,M)=1$, that is if $M$ is a positive root of the root system.