Integer angles

We have;
Angle sum of a regular polygon with $n$ sides $= (n-2)(180)$
Since, all angles of a regular polygon are equal hence;
Each angle = $\frac{(n-2)(180)}{n}$
The above fraction reduces to:
$180 - \frac {360}{n}$
Hence; the angles are integral for all factors of $360$ which are $24$ in number = $(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360$ .

So; the fraction is reduced to an integer for these $24$ values but we must note that $n$ cant be $1$ or $2$ since that wont make a polygon.

Hence; number of polygons $= 24 - 2 = \boxed {22}$

Note that all angles are in degrees

Number of divisors of a number 360 can be computed easily using the following theorem:

If $N = p_1 ^{q_1} p_2^{q_2} \ldots p_n ^ {q_n}$ , then $N$ has $d(N) = (q_1 +1) (q_2 +1) \cdots (q_n + 1)$ divisors.

In our case $360=2^3*3^2*5 \Rightarrow d(360)=(3+1)(2+1)(1+1)=24$ . For our purpose we exclude numbers $1 , 2$ from the list as polygons can not be formed with that many sides. The result is $24-2=\boxed{22}$ .

Each exterior angle of an regular polygon is $360/n$ degrees, and since interior angle + exterior angle = $180º$ , each interior angle is $180º - 360º/n$ . This is an integer only if $360/n$ is, so the problem reduces to finding the number of divisors of $360 = 2^3 \cdot 3^2 \cdot 5$ , which is $(3+1)(2+1)(1+1) = 24$ . However, since a polygon cannot have $1$ or $2$ sides, there are only $24 - 2 = \boxed{22}$ valid regular polygons.