The concept "opposite" of Cohen-Macaulayness Let $S = \mathbb k[x_1, \ldots, x_n]$ be a polynomial ring over a field $\mathbb k$ and let $I \subseteq S$ be a monomial ideal.
For a monomial ideal $J$, let $\#(J)$ be the smallest number of monomials that generate $J$.
A monomial quotient $S/I$ is called Cohen-Macaulay if the projective dimension of $S/I$ is equal to $\min\{\#(J): J \text{ is an irreducible component of }I\}$.
Is there a name for the "opposite" property that the projective dimension of $S/I$ is equal to $\max\{\#(J): J \text{ is an irreducible component of }I\}$? Are there prior work on these rings?
 A: I don't think this "opposite CM" property would be very interesting. Here is why. Let's first localize at a maximal ideal, so we can work over a local ring. For simplicity let me denote this local ring still by $S$.
1) Using $n$ as in the question, for any module $M$ over $S$ we have an equality
$$
{\rm proj.dim}\, M + {\rm depth}\, M = n
$$
2) Furthermore, for any module $M$ over $S$ we have an inequality
$$
{\rm depth}\, M \leq \dim M
$$
3) For an ideal $I$ generated by $r$ elements we have an inequality
$$
{\rm height}\, I \leq r.
$$
Based on your comment in response to my question in this case this is actually an equality, which implies that 
$$
\dim (S/I) = n -r.
$$
Putting 1)-3) together says that 
$$
{\rm proj.dim}\, (S/I) = n- {\rm depth}\, (S/I) \geq n-\dim(S/I) = r.
$$
In other words, you could rephrase the CM condition that the projective dimension of $S.I$ takes the smallest possible value this inequality allows. It also implies that all the irreducible components of $I$ are generated by the same number of elements.
The notion you are suggesting has no property that would influence the other irreducible components. I think you could do the following to have an example: Take an arbitrary ideal $I_0\subseteq S$ and let $r={\rm proj.dim}\,(S/I_0)$. Now let $\mathfrak p\subseteq S$ be a prime ideal of height $r$ that is generated by $r$ elements and does not contain any minimal primes of $I_0$ and let $I=\mathfrak p\cap I_0$. Then $S/I$ has your "opposite CM" condition, but there seems to be very little chance for some interesting behaviour.  
