1. Given data

• Least count of vernier caliper: $LC = \frac{1}{20N}\text{ cm}$
• One main scale division: $1\text{ MSD} = 1\text{ mm} = 0.1\text{ cm} = \frac{1}{10}\text{ cm}$

2. Use the vernier least count relation

For a direct vernier, $LC = 1\text{ MSD} - 1\text{ VSD}$

So, $\frac{1}{20N} = \frac{1}{10} - 1\text{ VSD}$

Hence, $1\text{ VSD} = \frac{1}{10} - \frac{1}{20N}$

Taking LCM, $1\text{ VSD} = \frac{2N-1}{20N}\text{ cm}$

3. Length of }N\text{ vernier divisions}

$N\text{ VSD} = N\times \frac{2N-1}{20N} = \frac{2N-1}{20}\text{ cm}$

4. Convert this into number of main scale divisions

Since $1\text{ MSD} = \frac{1}{10}\text{ cm}$

the number of main scale divisions equal to $N$ vernier divisions is $\frac{\frac{2N-1}{20}}{\frac{1}{10}} = \frac{2N-1}{2}$

5. Match with options

Thus, the number of main scale divisions that coincide with $N$ vernier divisions is $\boxed{\frac{2N-1}{2}}$

So the correct option is D.