One such result that springs to mind is that, if the finite group $G$ has a presentation with $r$ generators and $s$ relations, then the Schur Multiplier $M(G) = H_2(G)$ of $G$ can be generated by at most $s-r$ elements. So in particular $s \ge r$. (Of course you can prove that more directly - a finitely presented group with $s<r$ has infinite abelianization.)

We can deduce for example that a finite abelian group $G$ of rank $r$ requires at least $r(r+1)/2$ relations to present it, because $M(G)$ has rank $r(r-1)/2$. In this case the converse holds - the obvious presentation of $G$ has $r$ generators and $r(r+1)/2$ relations.

The converse does not hold in general. Swan constructed examples of finite solvable groups with trivial multiplier and arbitrarily large minimum $s-r$. But I believe it is still an open problem for finite $p$-groups. i.e. does every $d$-generator finite $p$-group have a presentation with $d$ generators and $d + {\rm rk}(M(G))$ relations?

Finite groups of defect 0 - i.e. $r=s$ have also been much studied. There are lots of 2-generator examples known, a few 3-generator examples and none requiring 4 or more generators.