What's the sum of all the positive integral divisors of $540$?

Assuming you want to know the sum of the divisors of 540, i.e. $$ \sigma_1(540) = \sum_{d \mid 540} d $$ then you can compute this by hand by either determining all divisors of 540, or you can do this a little more cleverly.

First of all, note that $540 = 2^2 \cdot 3^3 \cdot 5$. Now, one happy fact about the function $\sigma_1(n)$ is that if $n, m$ are relatively prime, then $\sigma_1(n \cdot m) = \sigma_1(n) \cdot \sigma_1(m)$. In particular, $$ \sigma_1(540) = \sigma_1(4)\sigma_1(27)\sigma_1(5) $$ and so you only need to determine these last ones. But for prime numbers, $\sigma_1(p^k)$ is easy to compute. It is $$ \sigma_1(p^k) = 1 + p + p^2 + \cdots + p^k = \frac{p^{k+1} - 1}{p-1} $$ which should let you compute the answer now.

Prime factorization of 540 is $2^2\cdot 3^3\cdot 5$. Also sum of divisors of 540 equals: $$\begin{align} \sigma\left(2^2\cdot 3^3\cdot 5\right) &=\frac {2^3-1}{2-1}\cdot\frac{3^4 - 1}{3-1}\cdot \frac{5^2 -1}{ 5-1}\\ &=7\cdot 40\cdot 6\\ &= 1680 \end{align}$$