If 16 women and 24 men can complete a work in 18 days, how many days will 24 men take to complete the same work, given that each woman takes twice as long as a man to complete the work?

The problem states that each woman takes twice as long as a man to complete the work. This means a man works twice as fast as a woman.
Let $M$ be the work rate of one man (work per day) and $W$ be the work rate of one woman (work per day).
Therefore, the work rate of a woman is half the work rate of a man:

$$W = \frac{1}{2} M$$

The combined work rate is $16W + 24M$. Substitute $W = \frac{1}{2} M$:

$$16\left(\frac{1}{2} M\right) + 24M = 8M + 24M = 32M$$

So, 16 women and 24 men together have a work rate equivalent to 32 men.

The group of 16 women and 24 men (equivalent to 32 men) completes the work in 18 days.
Total Work = (Equivalent men's work rate) $\times$ (Time taken)

$$\text{Total Work} = 32M \times 18 \text{ days}$$

Let $D$ be the number of days 24 men will take. The work rate of 24 men is $24M$.

$$\text{Total Work} = 24M \times D$$

Equate the total work from Step 3 and this expression:

$$32M \times 18 = 24M \times D$$

Divide both sides by $M$ (assuming $M \neq 0$):

$$32 \times 18 = 24 \times D$$

Solve for $D$:

$$D = \frac{32 \times 18}{24}$$ $$D = \frac{576}{24}$$ $$D = 24$$

Thus, 24 men will take 24 days to complete the work.

The number of days 24 men will take to complete the work is $\boxed{\text{24 days}}$.