Is there a (finite) ring with exactly 2 minimal left ideal and exactly 3 minimal right ideal? (rings are assumed to have identity)
Let $M$ be the set of all $3\times 2$ matrices over a field $F_2$ with usual addition. Let $P$ be the $2\times 3$ matrix consisting of $1$'s. Define a multiplication: $A*B= APB$. Then we get a ring (this is socalled Munn algebra) with required properties (a minimal left ideal is generated by a column). In such a way you can obtain a ring with $m$ minimal left ideals and $n$ minimal right ideals. Addendum (Thanks to Manny Reyes) $M$ doesn't contain an identity element, but we can join an identity $e$ to $M$, i.e. consider the ring $M_1$ of pairs $(a,me)$ ($a\in M, m\in F_2$) with multiplication $(a,me)*(b,ne)=(a*b+na+mb,mne)$. Then minimal ideals of $M_1$ are the same of $M$. 


Consider the ring $R = \Bbb{Z}_2<a,b \;  \; a^3 = b^2 = ba = a^2b = 0>$ ($a,b$ are noncommuting indeterminate). Then every element of $R$ is a linear combination of $1,a,b,a^2,ab$. Therefore $R = 32$, also $R$ is a local ring and satisfies the asked conditions. 

