Let N be a positive number. When the decimal point of N is moved 2 places to the left, the result is equal to N divided by 100 , or $\(\frac{N}{100}\)$.

According to the problem, this result is equal to $\(\frac{6}{N-1}\)$. We can set up the following equation:

$$
\(\frac{N}{100}=\frac{6}{N-1}\)
$$


To solve for N , we can cross-multiply:

$$
\(\begin{aligned}
& N(N-1)=6 \cdot 100 \\
& N^2-N=600
\end{aligned}\)
$$


Now, we can rearrange the equation into a standard quadratic form:

$$
\(N^2-N-600=0\)
$$


We can solve this quadratic equation by factoring. We are looking for two numbers that multiply to -600 and add up to -1 . The numbers are 24 and -25 .

So, we can factor the equation as:

$$
\((N+24)(N-25)=0\)
$$


This gives us two possible values for N :

$$
\(\begin{aligned}
& N+24=0 \Rightarrow N=-24 \\
& N-25=0 \Rightarrow N=25
\end{aligned}\)
$$


Since the problem states that N is a positive number, the only valid solution is $\(\mathbf{N}=\mathbf{2 5}\)$.
Let's check the answer:
- Moving the decimal point of 25 two places to the left gives 0.25.
- The expression $\(\frac{6}{N-1}\)$ becomes $\(\frac{6}{25-1}=\frac{6}{24}=0.25\)$.

Since both sides are equal, the value of $N$ is \(25 \).