How is it possible for the induced emf to take negative values in Faraday's Law of induction?

This is where it's a good time to "converse with the math". Let's look at the equation:

$$\mathcal{E} = \oint_\gamma \mathbf{E} \cdot d\mathbf{l}$$

which in this case equals $\mathcal{E}_\mathrm{ind}$. This is based on the work formula:

$$W = \int_\gamma \mathbf{F} \cdot d\mathbf{r}$$

Thus, what your question is, is essentially, asking "how can you have negative work". Looking at the above equation and recalling how a dot product behaves, there's only one way: $\mathbf{F}$ and the element of displacement $d\mathbf{r}$ must be aimed at cross purposes with each other. In the case of the integral for $\mathcal{E}_\mathrm{ind}$, the same goes only with $\mathbf{E}$ and $d\mathbf{l}$.

Hence, the answer to your question is: when $\mathbf{E}$ points opposite to $d\mathbf{l}$.

But what does that mean? Well, the key here is that we have to think a little more closely about the work formula. I believe what you are imagining it means is "the work done by the force as the force pulls the particle along with it". It actually is more general - in a work integral, the particle can be moved in any direction, including against the force. Of course, to make that motion happen in real life, you need to supply a source of contrary force, but that doesn't change the maths. This is why, say, in a more elementary example, you can talk of "negative work" done by the gravitational force when you lift an object off the floor.

(Why is it defined that way? Well, for one, because we often can't solve for the "real" trajectory the particle follows! If we stipulated that as a precondition, it would make work an extremely non-trivial concept!)

Where your mistake lies in, then, is in missing that. The displacement around the wire $d\mathbf{l}$ that we use to describe emf is not the displacement that necessarily occurs in reality to a real positive charge (after all, in many applications we aren't "really" dealing with positive charges anyways!).

Rather, it is a hypothetical one where we imagine that we grab a charge and move it around all the way through the circuit in a specific, fixed direction, and ask what the work - whether positive or negative - done by the electric force for that movement is. If we ask about the work done in the actual movements of charges, we will get a different answer, and yes, this one will always be positive, at least provided we don't get into looking at the situation in too-fine detail.

From an intuitive point of view, if you want to figure when the force is doing "positive" work and when it's doing "negative" work, imagine that you can feel the electric force tugging on the positive charge in your hand as you move it through the circuit. When you feel it helping you, i.e. the tug is with the motion of your hand, at that moment (i.e. that small increment $d\mathbf{l}$) the electric force is doing positive work, and you are doing negative work (to retard the charge, if you were to try and not naturally speed up your hand as you'd likely tend to, of course). When you feel it is fighting you, i.e. the tug is against the motion of your hand, the electric force is doing negative work, and you are doing positive work (to help it against the contrary pull). Total negative work, and hence negative emf, will be if, in moving the charge, you had to fight more often than flow.

The minus sign is necessary because the induced current must travel in a way such that the induced magnetic field produces a change in flux that is opposite to that of the original magnetic field

The phrase in bold is incorrect as it could be in the same direction as that of the original magnetic field if the magnitude of the original magnetic field is decreasing.

Consider a simple series circuit consisting of a cell of emf $\mathcal E_{\rm cell} $, a switch and an inductor $L$.

At a time $t=0$, when the current is the circuit $i=0$, the switch is closed.

$\mathcal {E_{\rm cell}} + \mathcal E_{\rm induced}=0\Rightarrow \mathcal {E_{\rm cell}} - L \frac {di}{dt}=0 \Rightarrow i = \frac{\mathcal E_{\rm cell}}{L}\,t$

After time $t$ the work done by the cell is $\displaystyle \int ^t_0 \mathcal E_{\rm cell} \,i\, dt = \frac{\mathcal E_{\rm cell}^2 \,t^2}{2L}= \frac 12 L i^2$ and this is minus the work done by the induced emf, $\displaystyle \int ^t_0 \mathcal E_{\rm induced} \,i\, dt=\int ^t_0 \left ( -L \frac {di}{dt}\right) \,i\, dt $, and represents the energy stored by the inductor.

Negative just signifies emf will such be induced that current induced will oppose change in magnetic field though it.It is not related to work it is just for above reason.