What kind of system would conserve momentum but not energy? In my physics book (chapter 9 - center of mass, linear momentum and collision), it is written

Momentum should not be confused with energy. In some cases momentum is conserved but energy is definitely not.

This is a bit confusing. So I want an example of a system where momentum is conserved but not energy.
 A: 
Momentum should not be confused with energy. In some cases momentum is
conserved but energy is definitely not.

This is indeed a confusing way of putting things. The first sentence is clear and unequivocal.
But the latter sentence should really read:

Momentum is always conserved but kinetic energy is not always conserved($^+$).

The latter part of this statement refers to energies that arise from non-conservative forces.

So I want an example of a system where momentum is conserved but not
energy.

Consider the so-called ballistic pendulum:

A bullet of mass $m$ and uniform velocity $v$ is fired into a soft bob of mass $M$. After collision bullet and bob do not separate and they reach height $h$. So the collision is entirely inelastic.
We know kinetic energy is not conserved here because for the bob and bullet to permanently unite the bullet must deform the bob, thereby doing work on it. There are other non-conservative energies at play here, so we can't apply simple conservation of energy.
But we can apply conservation of momentum, before and after the collision:
$$mv=(m+M)V$$
where $V$ is the velocity of bullet plus bob, immediately after the collision, so that:
$$V=\frac{m}{m+M}v$$
Post-collision, in the absence of air drag and other frictions, conservation of energy now does apply:
$$\Delta K=\Delta U$$
$$\frac12 (m+M)V^2=(m+M)gh$$
$$\frac12 \Big(\frac{m}{m+M}\Big)^2(m+M)v^2=(m+M)gh$$
So that:
$$h=\frac{1}{2g} \Big(\frac{m}{m+M}\Big)^2 v^2$$

So how much energy was 'lost' (not conserved)? Just look at the TOTAL energy balance:
$$\frac12 mv^2=(m+M)gh+E_{lost}$$
With everything else known, $E_{lost}$ can easily be calculated. It calculates to:
$$E_{lost}=\frac12 \frac{mM}{m+M}v^2$$
It's basically the kinetic energy that was lost during the inelastic collision.
If we express it as a ratio:
$$\frac{E_{lost}}{\frac12 mv^2}=\frac{M}{m+M}$$
So if $M\gg m$ then the ratio becomes $1$: ALL of the bullet's kinetic energy is 'lost'.
($^+$) Re. conservation of energy, please note what I wrote in the comments:

If you consider ALL energies, including those stemming from
non-conservative forces, then energy is ALWAYS conserved.

But as it happens, in some cases several energies are very hard to know, in the example above e.g. the work done by the bullet on the bob, noise from the collision, heat from friction etc.
A: It should have said that ''kinetic energy'' is not always conserved.
When you throw some sticky dough at a wall the kinetic energy is converted completely to heat and sound energy and the dough sticks to the wall. Momentum is conserved though. When you throw the dough the dough went one way and the entire Earth went the other. When the dough stuck to the wall both the dough and the Earth stopped moving. Kinetic energy was not conserved. Energy was conserved.
If you throw a ball at the wall it will bounce back. Kinetic energy was not all converted to heat and sound but some of it was. If it's a really really bouncy ball then we say that kinetic energy is conserved. But in reality no such ball exists. Some energy is always converted to heat on contact.
Energy is always conserved but kinetic energy is not always conserved.
Momentum is always conserved.
A: Perhaps the simplest example is two identical objects moving toward each other with the same speed and sticking together at rest after colliding. Initial momentum = 0. Final momentum = 0. So momentum is conserved. Initial kinetic energy is nonzero but final kinetic energy is 0.
