How many numbers do we need?

The primes less than 100 are:

$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97$

Any prime less than 50 has a multiple between 53 and 97. Therefore, the "shortest" $p$ which is divisible by all the primes less than 100 is $p = 53 \times 54 \times 55 \times \cdots \times 97$ , where the number of consecutive natural numbers is $n = 97-53+1 = \boxed{45}$ .

No prime bigger than $50$ has any multiple other than itself less than $100$ . This means we need all of the primes (from $53$ to $97$ ) to be included in the list.

Every prime less than $50$ has at least one multiple in the range $53$ to $97$ , so this is actually enough; there are $97-53+1=\boxed{45}$ consecutive numbers in this list.

With the Sieve of Eratosthenest we can easily find all prime numbers less than 100.

(1) 2; 3; 5; 7; 11; 13; 17; 19; 23; 29; 31; 37; 41; 43; 47;

(2) 53; 59; 61; 67; 71; 73; 79; 83; 89; 97

The lowest multiple of 53 is 106, therefore every prime of the (2) row has to be one of those consecutive numbers. Now we have 45 consecutive numbers. If we have $j$ consecutive numbers, then there is at least one of them, which is devisible by $j$ . Therefore if we take 45 consecutive numbers from 53 to 97, $p$ is devisible by all numbers of the (1) row except 47, but $2 \cdot 47 = 94$ . Our $p$ is devisible by all prime numbers less than 100.

Thus $n = \boxed { 45 }$