1. Divisibility by $55$

A number is divisible by $55$ iff it is divisible by both $5$ and $11$.

• Divisible by $5$ $\Rightarrow$ last digit must be $5$ (since available digits are $2,3,4,5,6$).
• Divisible by $11$ $\Rightarrow$ difference between sum of digits in odd and even places is a multiple of $11$.

So every required number must end in $5$.

2. Possible number of digits

We need numbers between $1012$ and $23421$.

Using digits $2,3,4,5,6$ without repetition:

• $4$-digit numbers are possible.
• $5$-digit numbers are possible.

We count both, subject to the range.

3. Case 1: 4-digit numbers

Let the number be $abcd$, with $d=5$. So the number is of the form $abc5$, where $a,b,c$ are chosen from $\{2,3,4,6\}$ without repetition.

For divisibility by $11$:

Since the digits are small, only $0$ is possible. Thus,

Now check permutations of $\{2,3,4,6\}$ taken $3$ at a time.

Equivalently,

Try each possible $b$:

• If $b=2$, then $a+c=7$. From remaining digits $\{3,4,6\}$, only $3+4=7$ works. Hence numbers: $3245,\ 4235$.

• If $b=3$, then $a+c=8$. From remaining digits $\{2,4,6\}$, only $2+6=8$ works. Hence numbers: $2365,\ 6325$.

• If $b=4$, then $a+c=9$. From remaining digits $\{2,3,6\}$, only $3+6=9$ works. Hence numbers: $3465,\ 6435$.

• If $b=6$, then $a+c=11$. From remaining digits $\{2,3,4\}$, no pair sums to $11$.

So total $4$-digit numbers $=6$.

All of these are greater than $1012$, so all are valid.

4. Case 2: 5-digit numbers

Let the number be $abcde$, with $e=5$. So it is of the form $abcd5$, where $a,b,c,d$ are a permutation of $2,3,4,6$.

Divisibility by $11$ gives:

Now,

So

Hence,

Thus we need

But the left side is even, so it cannot be $\pm 11$. Therefore,

So $a,c$ must be $2,3$ in some order, and then $b,d$ are $4,6$ in some order.

Thus possible 5-digit numbers are: $24365,\ 26345,\ 34265,\ 36245.$

Now apply the range condition: number must be less than $23421$.

Among these,

• $24365 > 23421$
• $26345 > 23421$
• $34265 > 23421$
• $36245 > 23421$

So none of the 5-digit numbers are allowed.

5. Total count

Only the $4$-digit numbers work, and there are $6$ of them.

Therefore, the required number of natural numbers is

6. Comparison with stored answer

Stored correct answer: $6$

Our derived answer is also $6$, so they agree.