Problem 1223

1. Define the Problem:
We need to determine which natural numbers cannot be written as the sum of consecutive natural numbers and prove such impossibility.

2. Claim 1:
If $n$ has an odd divisor $k > 1$, then $n$ is a beautiful number (i.e., it can be written as the sum of consecutive natural numbers).

Proof:
- If $n$ is an odd number, then we can write $n = a + (a + 1)$ with $a = \frac{n-1}{2}$.
- If $n$ is an even number, then write $n = 2^s \cdot t$ with $t > 1$ being an odd number.
- Let $b = \frac{t-1}{2}$.
- If $2^s \ge b$, then we can write $n = i + (i+1) + \ldots + (i+t-1) = t(i+b)$ with $i = 2^s - b$.
- If $b > 2^s$, then we can write $n = [b - (2^s - 1)] + [b - (2^s - 2)] + \ldots + b + (b+1) + \ldots + (b + 2^s)$.

3. Claim 2:
If $n$ is a beautiful number, then there exists an odd natural number $k > 1$ such that $k \mid n$.

Proof:
- Since $n$ is a beautiful number, there exist $i, h \in \mathbb{N}, h > 1$ such that
$$n = i + (i+1) + (i+2) + \ldots + (i+h-1) = h \left( i + \frac{h-1}{2} \right)$$
- If $h$ is odd, then $\frac{h-1}{2} \in \mathbb{N}$, so $h \mid n$ and we are done.
- If $h$ is even, then
$$n = \frac{h}{2} (2i + h - 1)$$
so $2i + h - 1$ is odd and $2i + h - 1 \mid n$, we are done.

4. Conclusion:
From Claims 1 and 2, we have shown that $n$ is beautiful if and only if $n$ has an odd divisor greater than 1. Therefore, the numbers that cannot be written as the sum of consecutive natural numbers are those that do not have any odd divisors greater than 1. These numbers are the powers of 2.

Hence, the numbers that cannot be written as the sum of consecutive natural numbers are $2^s$ with $s \in \mathbb{N}$, for example, $1, 2, 4, 8, \ldots$.

The final answer is $\boxed{ 2^s }$ with $s \in \mathbb{N}$.