Find the smallest value of $n$ such that the sum of the first $n$ natural numbers is greater than 1,000.

We need: $1 + 2 + 3 + \dots + n > 1000$

Using the sum formula:

$$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$$

$$\frac{n(n+1)}{2} > 1000$$

$$n(n + 1) > 2000$$

Check nearby values:

$$44 \times 45 = 1980$$

$$45 \times 46 = 2070$$

So, $$\frac{44 \times 45}{2} = 990$$

$$\frac{45 \times 46}{2} = 1035$$

Hence, the smallest value of $n$ is $45$.