A scaling by $N$ of the DFT result is required to have the same amplitude as the time domain signal. Consider the simple case of a sinusoid where the sampling rate is an integer multiple of the frequency (to avoid spectral leakage for a simple example):

$$x[n] = A\cos(2 \pi (f/f_s) n)$$

Where $A$ is the real magnitude, and $f_s$ is the sampling rate as an integer multiple of $f$.

Using Euler's relationship we know this is:

$$x[n] = \cos(\omega_n n) = \frac{A}{2}e^{j\omega_n n} +  \frac{A}{2}e^{-j\omega_nn}$$

Where   $\omega_n =  2 \pi (f/f_s)$.

Every component in the DFT result is the coefficient for a $e^{j\omega n}$ in the time domain, so the DFT of a sinusoid with the corrected magnitude should have two components each with magnitude $A/2$ per the formula shown above.

Observe in the DFT formula given as:

$$X(k) = \sum_{n=0}^{N-1}x[n]e^{-j2\pi n k/N}$$

When $2\pi n k/N = \omega_n$, the product in the summation will be $A/2$ and so the sum will grow to $NA/2$, so scaling by $N$ results in the matching amplitude of the coefficient of the $e^{j\omega t}$ time domain waveform.

Further, since the magnitude for $\frac{A}{2}e^{\omega_n t}$ is constant the power is simply $ (\frac{A}{2})^2$ and not divided by $2$ as we would do with the sinusoid. And all bins sum in power so that the total power of $x[n]$ correctly equals $ (\frac{A}{2})^2 + (\frac{A}{2})^2 = 2(\frac{A}{2})^2 = \frac{A^2}{2}$ and we happily get that factor of 2 for the sinusoid we are familiar with. 

For single tones and narrow band signals that are less than the resolution bandwidth of the DFT bin, there is no effect on resolution bandwidth as the power for the signal has no distribution (meaning the power of the signal occupies a bandwidth that is 0 wide, so no matter how tight or loose we make the resolution bandwidth on a spectrum analyzer like the DFT we will get the same magnitude result, when properly scaled as I demonstrated). It is when we get into waveforms with bandwidth (and noise in general) that we must be careful about the effects of resolution bandwidth, windowing, etc in using the DFT for accurate power spectral density measurements. I address this in these other posts here on StackExchange: