What is the technical definition of an observation/measurement?
A QM measurement is essentially a filter. Observables are represented by operators $\smash {\hat O}$, states or wave functions by (linear superpositions of) eigenstates of these operators, $|\,\psi_1\rangle, |\,\psi_2\rangle, \ldots$. In a measurement, you apply a projection operator $P_n$ to your state, and check if there is a non-zero component left. You ascertain you yourself that the system is now in the eigenstate $n$. Experimentally, you often send particles through a "filter", and check if something is left. Think of the Stern-Gerlach experiment. Particles that come out in the upper ray have spin $S_z = +\hbar/2$. We say we have measured their spin, but we have actually $prepared$ their spin. Their state now fulfils $\smash{\hat S} \,|\,\psi\rangle = +\hbar/2 \,|\,\psi\rangle$, so it is the spin-up eigenstate of $\smash{\hat S}$. This is physical and works even if no one is around.
If I look at a QM system, it will collapse. But how is that any different from a bunch of matter "looking" at the same system?
Can the system tell the difference between a person's eyes and the bunch of matter?
There are two different things going on, knowledge update (subjective), and decoherence (objective).
First the objective part: If you have a quantum system by itself, it's wave function will evolve unitarily, like a spherical wave for example. If you put it in a physical environment, it will have many interactions with the environment, and its behavior will approach the classical limit.
Think of the Mott experiment for a very simple example: Your particle may start as a spherical wave, but once it hits a particle, it will be localized, and have a definite momentum (within $\Delta p \,\Delta x \geq \hbar/2$). That's part of the definition of "hits a particle". The evolution will then continue from there, and it is very improbable that the particle has the next collision in the other half of the chamber. Rather, it will follow its classical track.
Now the subjective part: If you look at a system, and recognize that it has certain properties (e.g. is in a certain eigenstate), you update your knowledge and use a new expression for the system. This is simple, and not magical at all. There is no change in the physical system in this part; a different observer could have different knowledge and thus a different expression. This subjective uncertainty is described by density matrices.
Sidenote on density matrices:
A density matrix says you think the system is with probability $p_1$ in the pure state $|\,\psi_1\rangle$, with probability $p_2$ in the pure state $|\,\psi_2\rangle$, and so on. (A pure state is one of the states defined above and can be a superposition of eigenstates, where as a mixed state is one given by a density matrix.)
Pure states are objective, if I have a bunch of spin-up particles from my Stern-Gerlach experiment, my colleague will have to agree that they are spin-up, no matter what. They all go in his experiment to the top, too. If I have a bunch of undetermined-spin particles, $$|\,\psi\rangle_\mathrm{undet.} = \frac{1}{\sqrt{2}} (|\,\psi_\uparrow\rangle + |\,\psi_\downarrow\rangle)\,,$$
they will turn out 50/50, for both of us.
Mixed states are different. My particles could be all spin-up, but I don't know that. Someone else does, and he uses a different state to describe them (e.g. see this question). If I see them fly through a magnetic field, I can recognize their behavior, and use a new state, too.
And note that a mixed state of 50% $|\,\psi_\uparrow\rangle$ and 50% $|\,\psi_\uparrow\rangle$ is not the same as the pure state $|\,\psi\rangle_\mathrm{undet.}$ defined above.
If not, how can the system remain QM?
Technically, it remains QM all the time (because classical behavior is a limit of QM, and physics always has to obey QM uncertainties). Of course, that's not what you mean. If a system is to stay in a nice, clean quantum state for a prolonged time, it helps that it is isolated. If you have some interaction with the environment, it will not neccessarily completely decohere and become classical, but a perfect QM description will become impractically complicated, as you would have to take the environment and the apparatus into account quantum mechanically.