Let $n$ be a natural number. We want to count the number of $n < 1000$ that can be written as the sum of two or more consecutive natural numbers.

Let the consecutive numbers be $a, a+1, \ldots, a+k-1$ for $k \geq 2$. Their sum is:
$$S = a + (a+1) + \cdots + (a+k-1) = k a + \frac{k(k-1)}{2}$$
So $n = k a + \frac{k(k-1)}{2}$ for some $k \geq 2$ and $a \geq 1$.

Solving for $a$:
$$ka = n - \frac{k(k-1)}{2} \implies a = \frac{n - \frac{k(k-1)}{2}}{k}$$
We require $a \geq 1$ and $a$ integer, so $n - \frac{k(k-1)}{2}$ must be divisible by $k$ and $n \geq \frac{k(k+1)}{2}$.

But the key is: Which $n$ cannot be written in this way?

It is a well-known result that the numbers that cannot be written as the sum of two or more consecutive natural numbers are the powers of $2$ (i.e., $1, 2, 4, 8, 16, \ldots$).

Proof:
Let $n$ be a natural number. $n$ can be written as the sum of $k$ consecutive natural numbers if and only if there exists $k \geq 2$ such that $n - \frac{k(k-1)}{2}$ is divisible by $k$ and $n \geq \frac{k(k+1)}{2}$.

But the only numbers that cannot be written in this way are the powers of $2$.

So, the number of special numbers less than $1000$ is $999$ minus the number of powers of $2$ less than $1000$.

The powers of $2$ less than $1000$ are:
$1, 2, 4, 8, 16, 32, 64, 128, 256, 512$

$2^{0} = 1$
$2^{1} = 2$
$2^{2} = 4$
$2^{3} = 8$
$2^{4} = 16$
$2^{5} = 32$
$2^{6} = 64$
$2^{7} = 128$
$2^{8} = 256$
$2^{9} = 512$
$2^{10} = 1024 > 1000$

So there are $10$ powers of $2$ less than $1000$.

Therefore, the answer is:
$$999 - 10 = 989$$

Answer: There are $989$ special natural numbers less than $1000$.