**Solution:**

To find the number of positive integers co-prime to 480 and less than it, we will use the Euler's totient function, denoted by $\phi(n)$. The Euler's totient function gives the number of positive integers less than or equal to n that are relatively prime to n.

**Using Euler's totient function:**

We have to find the value of $\phi(480)$. We can find it using the formula:

$$\phi(n) = n\left(1 - \frac{1}{p_1}\right)\left(1 - \frac{1}{p_2}\right)\dots\left(1 - \frac{1}{p_k}\right)$$

where $p_1, p_2, \dots, p_k$ are the distinct prime factors of n.

Using this formula, we can write:

$$\phi(480) = 480\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{5}\right)$$
$$= 480 \times \frac{1}{2} \times \frac{2}{3} \times \frac{4}{5} = 96$$

Therefore, there are 96 positive integers less than 480 that are co-prime to 480.

**Finding the Sum:**

To find the sum of all those numbers, we can use the formula for the sum of first n natural numbers:

$$S_n = \frac{n(n+1)}{2}$$

We can divide the numbers co-prime to 480 into two sets:

1. Set A: Numbers that are co-prime to 480 and are odd.
2. Set B: Numbers that are co-prime to 480 and are even.

**Finding the sum of set A:**

The numbers that are co-prime to 480 and are odd form an arithmetic sequence with a common difference of 2, starting from 1. The last term of this sequence is the largest odd number less than 480, which is 479. Therefore, the number of terms in this sequence is:

$$n_A = \frac{479-1}{2} + 1 = 240$$

Using the formula for the sum of first n odd numbers, we can find the sum of set A:

$$S_A = n_A^2 = 240^2 = 57600$$

**Finding the sum of set B:**

The numbers that are co-prime to 480 and are even form an arithmetic sequence with a common difference of 4, starting from 2. The last term of this sequence is the largest even number less than 480, which is 478. Therefore, the number of terms in this sequence is:

$$n_B = \frac{478-2}{4} + 1 = 120$$

Using the formula for the sum of first n even numbers, we can find the sum of set B:

$$S_B = n_B(n_B+1) = 120 \times 121 = 14520$$

**Finding the total sum:**

The sum of all the numbers that are co-prime to 480 is:

$$S = S_A + S_B = 57600 + 14520 = 72120$$

128 and 30720