"Life is really worth living in a Noetherian ring $R$ when all the local rings have the property that every s.o.p. is an R-sequence. Such a ring is called Cohen–Macaulay (C–M for short).": Hochster, "Some applications of the Frobenius in characteristic 0", 1978.
Section 3 of that paper is devoted to explaining what it "really means" to be Cohen–Macaulay. It begins with a long subsection on invariant theory, but then gets to some algebraic geometry that will interest you.
In particular, he points out that if $R$ is a standard graded algebra over a field, then it is a module-finite algebra over a polynomial subring $S$, and that $R$ is Cohen–Macaulay if and only if it is free as an $S$-module. Equivalently, the scheme-theoretic fibers of the finite morphism $\operatorname{Spec} R \to \operatorname{Spec} S$ all have the same length.
At the end of section 3, Hochster explains that the CM condition is exactly what is required to make intersection multiplicity "work correctly": If $X$ and $Y$ are CM, then you can compute the intersection multiplicity of $X$ and $Y$ without all those higher $\operatorname{Tor}$s that Serre had to add to the definition.
He gives lots of examples and explains "where Cohen–Macaulayness comes from" (or doesn't) in each one. The whole thing is eminently readable and highly recommended.
Dear Andrea: Hartshorne was right, but we need to do some work. Let $\mu(I)$ be the minimal number of generators of $I$, and $\mu_h(I)$ be the minimal number of a homogenous system of generators of $I$. Let $R=k[x_1,\dots,x_n]$ and $\mathfrak m=(x_1,\dots,x_n)$. Suppose $\mu_h(I)=m$ and $f_1,\dots, f_m$ is a minimal homogenous set of generators. At this point we switch to the local ring $A=R_{\mathfrak m}$ (the reason: it is easier to do linear algebra over local rings, as anything not in $\mathfrak m$ is now invertible). It will not affect anything since $I\subset\mathfrak m$.
Construct a surjective map $F_0 = \bigoplus_{i=1}^m A(-\deg \ f_i) \to I \to 0$ and let $K$ be the kernel. We claim that $K \subset\mathfrak mF_0$. If not, then one can find an element $(a_1,...,a_m) \in K$ such that $\sum a_if_i=0$ and $a_1$, say, has a degree $0$ term $u_1\neq 0$. By considering terms of same degree in the sum one sees that there are $b_i$s such that:$$u_1f_1 = \sum_{i=2}^m b_if_i$$
so the system is not minimal, as $u_1 \in k$, contradiction.
Now tensoring the sequence $$ 0 \to K \to F_0 \to I \to 0$$ with $k=A/\mathfrak m$. By the claim $K\subset\mathfrak mF_0$, so $F\otimes k \cong I\otimes k$. It follows that $m= \operatorname{rank} F_0 = \dim_k(I\otimes k)$. But over a local ring, the last term is exactly $\mu(I)$, and we are done.
Best Answer
If $X\subset \mathbb P^N$ is an l.c.i projective subvariety that linearly spans $\mathbb P^N$, and if $p\in\mathbb P^n\setminus X$ is a point s.t. the projection of $X$ from $p$ into $\mathbb P^{N-1}$ is an isomorphism onto its image $X'$, then $X'$ will not be a complete intersection (since the linear system of its hyperplane section is not complete).
To obtain such an example, suppose that $Y\subset\mathbb P^N$ is a smooth surface not lying in a hyperplane and that $N$ is big enough ($N\ge6$ will suffice). If characteristic is zero, there exists a quadric $Q\subset\mathbb P^N$ that is tangent to $Y$ at exactly one point $y\in Y$. Now if $X=Y\cap Q$ and $P$ is a generic point of $\mathbb P^N$ (in particular, $p$ should not lie in the tangent space $T_yY$), then the projection $\pi_p\colon X\to \mathbb P^{N_1}$ is an isomorphism onto ints image, and $\pi_p(X)\subset\mathbb P^{N-1}$ is an l.c.i that is not a c.i.