$\textbf{Problem 1.}$ Alejandra is going to distribute candies to several children. He gives the first one a candy, the second gets two, the third gets twice as many candies as he gave the second and so on, if Alejandra has $2007$ candies, what is the minimum number of candies that is missing to be able to distribute them exactly? and how many children would you give candy?
1. **Determine the number of candies given to each child:**
   - The first child receives \(1\) candy.
   - The second child receives \(2\) candies.
   - The third child receives \(2 \times 2 = 4\) candies.
   - The fourth child receives \(2 \times 4 = 8\) candies.
   - In general, the \(n\)-th child receives \(2^{(n-1)}\) candies.

2. **Calculate the total number of candies needed for \(n\) children:**
   - The total number of candies \(S_n\) given to \(n\) children is the sum of a geometric series:
     \[
     S_n = \sum_{i=0}^{n-1} 2^i = 2^0 + 2^1 + 2^2 + \cdots + 2^{(n-1)}
     \]
   - The sum of this geometric series is:
     \[
     S_n = \frac{2^n - 1}{2 - 1} = 2^n - 1
     \]

3. **Find the smallest \(n\) such that \(S_n \geq 2007\):**
   - We need to find the smallest \(n\) such that \(2^n - 1 \geq 2007\).
   - Solving for \(n\):
     \[
     2^n - 1 \geq 2007 \implies 2^n \geq 2008
     \]
   - The smallest power of 2 greater than or equal to 2008 is \(2^{11} = 2048\).

4. **Calculate the number of candies missing:**
   - The total number of candies needed for \(11\) children is:
     \[
     S_{11} = 2^{11} - 1 = 2048 - 1 = 2047
     \]
   - The number of candies Alejandra has is 2007.
   - The number of candies missing is:
     \[
     2047 - 2007 = 40
     \]

5. **Determine the number of children Alejandra can give candies to:**
   - Since \(2^{11} = 2048\) is the smallest power of 2 greater than 2007, Alejandra can give candies to \(11\) children.

The final answer is \( \boxed{ 40 } \) candies are missing, and Alejandra can give candies to \(11\) children.