On a board, Ana writes $a$ different integers, while Ben writes $b$ different integers. Then, Ana adds each of her numbers with with each of Ben’s numbers and she obtains $c$ different integers. On the other hand, Ben substracts each of his numbers from each of Ana’s numbers and he gets $d$ different integers.

For each integer $n$ , let $f(n)$ be the number of ways that $n$ may be written as sum of one number of Ana and one number of Ben.

[i]a)[/i] Show that there exist an integer $n$ such that,
$$f(n)\geq\frac{ab}{c}.$$
[i]b)[/i] Does there exist an integer $n$ such that,
$$f(n)\geq\frac{ab}{d}?$$

[i]Proposed by Besfort Shala, Kosovo[/i]
a) Let \( A \) be the set of integers written by Ana and \( B \) be the set of integers written by Ben. We define the set \( A + B \) as follows:
\[ A + B = \{ x + y \mid x \in A, y \in B \} \]
Similarly, we define \( A - B \) as:
\[ A - B = \{ x - y \mid x \in A, y \in B \} \]

Given that \( |A| = a \) and \( |B| = b \), the total number of pairs \((x, y)\) where \( x \in A \) and \( y \in B \) is \( ab \).

The function \( f(n) \) represents the number of ways \( n \) can be written as the sum of one number from \( A \) and one number from \( B \). Therefore, \( f(n) \) counts the number of pairs \((x, y)\) such that \( x + y = n \).

By the Pigeonhole Principle, if we distribute \( ab \) pairs into \( c \) different sums (since there are \( c \) different integers in \( A + B \)), there must be at least one sum \( n \) that appears at least:
\[ \left\lceil \frac{ab}{c} \right\rceil \]
times. This is because if each sum appeared fewer than \( \frac{ab}{c} \) times, the total number of pairs would be less than \( ab \), which is a contradiction.

Thus, there exists an integer \( n \) such that:
\[ f(n) \geq \frac{ab}{c} \]

The final answer is \( \boxed{ f(n) \geq \frac{ab}{c} } \)