Find the sum of all natural numbers between 250 and 550 which are divisible by 5.

The natural numbers must be between 250 and 550, and divisible by 5.
The first number greater than 250 that is divisible by 5 is 255. So, the first term $a = 255$.
The last number less than 550 that is divisible by 5 is 545. So, the last term $l = 545$.
Since the numbers are divisible by 5, the common difference $d = 5$.

We use the formula for the $n$-th term of an arithmetic progression: $l = a + (n-1)d$.

Subtract 255 from both sides:

Divide by 5:

Add 1 to both sides:

There are 59 numbers in the sequence.

We use the formula for the sum of an arithmetic progression: $S_n = \frac{n}{2}(a + l)$.

The sum of all natural numbers between 250 and 550 which are divisible by 5 is $\boxed{23600}$.