Reason for protonation of amino group in amino acids at near neutral pH
Prerequisite knowledge

A proton is deprotonated when the $\mathrm{pH}$ exceeds the $\mathrm pK_\mathrm a$ of the proton.


Analysis
There are 3 different forms that an amino acid can exist in. For the sake of simplicity of analysis, I am taking the simplest amino acid - glycine.
$$\underset{\mathrm{glycine}}{\ce{HOOC-CH2-NH2}}$$
Continuing to state the conditions of analysis, let's take a highly acidic medium to begin with and observe the changes that take place on increasing $\ce{pH}$.
Case 1: pH of solution &lt; 2.34
In such a scenario, the $\mathrm{pH}$ is lower than the $\mathrm p K_\mathrm a$ of the acid group and lower than the $\mathrm p K_\mathrm a$ of the amino group. This means that both the amino group and acid groups are protonated. The amino acid is positively charged at this $\mathrm {pH}$.
$$\underset{\mathrm{glycine\,(pH \,&lt;\,2.34)}}{\ce{HOOC-CH2-NH3+}}$$
Now, we increase the $\ce{pH}$ to a value greater than $2.34$ but less than 9.6  Specialty of these numbers? We'll see at the end.
Case 2: pH of solution greater than 2.34 but less than 9.6
In such a scenario, the $\mathrm{pH}$ is higher than the $\mathrm p K_\mathrm a$ of the acid group but lower than the $\mathrm p K_\mathrm a$ of the amino group. This means that the amino group remains protonated and the acid group is now deprotonated.
$$\underset{\mathrm{glycine\,(2.34\,&lt;\,pH \,&lt;\,9.6)}}{\ce{^-OOC-CH2-NH3+}}$$
Case 3: pH of solution greater than 9.6
In this last case, the $\mathrm{pH}$ is higher than both the $\mathrm pK_\mathrm a$s of the amino group and the acid group. This means that both groups are now deprotonated. The amino acid is now negatively charged.
$$\underset{\mathrm{glycine\,(pH \,&gt;\,9.6)}}{\ce{^-OOC-CH2-NH2}}$$

Observations and Conclusion
As you noticed, there are some specific numbers that are so important they are mentioned. They are the $\mathrm p K_\mathrm a$ values of the acid group ($\mathrm pK_{\mathrm a_1}$) and the amino group ($\mathrm pK_\mathrm {a_2}$). There is one more important number and that is the exact point of net average neutrality. It is known as the isoelectric point.
$$\mathrm {pI} = \frac{\mathrm pK_\mathrm {a_1} + \mathrm pK_\mathrm {a_2}}{2}$$