Number of discrete samples required for the longest wavelength
If it is a real signal, as you are doing, then the negative spectrum will be the complex conjugate of the positive spectrum. Since there are N samples in frequency given N samples in time, and the frequency axis extends from 0 to N-1 where 0 represents DC and N represents the sampling rate (with the upper half of this from N/2 to N-1 equally representing the negative frequency axis) then yes in this case you need 2N samples to represent the spectrum without aliasing; and yes this is explained by the Nyquist-Shannon sampling theorem: https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem

However if you used an equivalent complex signal, then N samples would be sufficient.

$$x[n]=e^{j(30 \cdot 2\pi n/N)} + e^{j(60 \cdot 2\pi n/N)}  + e^{j(120 \cdot 2\pi n/N)} \\  + e^{j(260 \cdot 2\pi n/N)} + e^{j(330 \cdot 2\pi n/N)}$$