The sum of a few natural numbers.

The only such natural numbers are $1$ and $2$ , with sum $\boxed3$ .

There are no natural numbers smaller than $1$ , so let's assume $n>1$ . If all numbers less than or equal to $n$ divide $n$ , then in particular $n-1$ divides $n$ . But $\frac{n}{n-1}=1+\frac{1}{n-1}$ , and for this to be a whole number, we need $n-1$ to divide $1$ . The only factor of $1$ is $1$ ; so the only other solution is $n=2$ , and we're done.

For the first 2 natural numbers ( $1$ and $2$ ) this is clearly true. For all other numbers:

If a natural number $n$ is divisible by all smaller natural numbers. $n$ is divisible by all primes less than $n$ .

The next number after $2$ , which could be such a number is $4$ . But the product of all primes less than $4$ is $2 \cdot 3 = 6$ . Thus the next possible number after $2$ is $6$ . But $6 \pm 1$ is a prime.

This is true, because : $lcm( x ,x\pm 1) =1$ .

Generally the product of the first $p$ primes ( ${ p }_{ sum }={ p }_{ 1 }\cdot { p }_{ 2 }\cdot \dots \cdot { p }_{ x-1 }\cdot { p }_{ x }$ ) is not divisible by all primes less than ${ p }_{ sum }$ because ${ p }_{ sum }-1$ is a prime and is not included in ${ p }_{ sum }$ .

Thus $1$ and $2$ are the only natural numbers for which this holds true and the sum is $3$ .