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.

largest numberof generators of any irreducible component of $I$? $\endgroup$ – Sándor Kovács Sep 11 '16 at 21:48