Designing a Chebyshev filter: am I getting the order wrong?
Your calculations are right, only your presumption is off: you are required to have a 3 dB gain, which means passive filters are out of the question, which also means you are no longer restricted to odd orders, because only passive filters suffer from attenuation at DC. If it's an active filter, the DC will have a 3 dB amplification, the ripple will contribute with an additional 3 dB, and the attenuation is considered (unless specified) at max peak minus ripple, thus +3 dB. The solution requires you to prove that n=4, so you got it right when you calculated that \$n\ge3.85\$. Congratulations!
A quick check proves it -- the amplification is 3 dB (cursor 1 is at DC), the same cursor shows that fc is at 1 kHz and has the same amplification, 3 dB, and the 2nd cursor shows -60 dB attenuation at 4 kHz which, added with the 3 dB in the passband, make the almost 63 dB difference that is shown:


There isn't much else that can be said, except to clear out the  possible confusion regarding the difference between the gain of the passive and active filters.
For one, if it's passive, the DC response can never be 0 dB or higher (well, technically it can be, but that's more of an exception), and that makes sense since there are an input and an output impedance which, considering the filter lossless, would mean that the highest output voltage would be the one given by \$\frac{Z_{out}}{Z_{in}+Z_{out}}\$.
However, in the case of the filters with passband ripple, such as Chebyshev, there is a peculiarity at even orders, because the transfer function causes the alternances of the passband ripple to start at the DC value of \$1-\delta_p\$ (relative to the maximum in the passband). This also means that the output impedance must always be less than the input. There is a whole passive filter synthesis that I won't repeat here (@jonk also has a very nice document linked in the comments), but I'll say that the load needs to be less than or equal to \$\small\tanh\left[\mathrm{asinh}\left(\frac{1}{\epsilon_p}\right)\right]^2\$ (relative to \$Z_{in}\$), if the load is fed by a series inductor, or the reciprocal if the load is fed by a parallel capacitor (a sweep for orders from 2 to 7):

Nothing will blow up if you'll use a different value, but the response will not be Chebyshev, anymore. For example, calculating for, and using unity I/O impedances:

Active filters can have any DC level and, thus, they can compensate for the loss of the passive filters. The attenuation at the corner frequency now will shift to \$1+\delta_p\$ for even orders, meaning the ripples will be over unity, and that the problem of compensating for the ripples is now inverted:

This is the case for your problem, where you need an active filter in order to have an over-unity gain, and thus choose an even order. If your teacher told you that you should compare the results with an odd order, then you should account for the \$1+\delta_p\$.
I'll add a final note. Suppose you actually end up with \$n=4.01\$ (for your pronlem, as shown, it's not the case), but if you do, then that requires a choice:

if the design is strict, that is, the minimum attenuation must be preserved at all costs, then there must be a rounding up
if, however, the tolerances account for a slight amplification that can result in rounding down (there is, after all, only a tiny fraction in there that's extra), then the lesse order can be chosen.

Mathematically, and since this is an exercise, you would need to use the rounding up (unless your teacher says it can be otherwise). To what order? That, again, depends on the constraints. Here, it would have been 5, since there is no passive filter.