At a party there are $n$ women and $n$ men. Each woman likes $r$ of the men, and each man likes $s$ of then women. For which $r$ and $s$ must there be a man and a woman who like each other?
1. **Setup the problem:**
   We have \( n \) women and \( n \) men. Each woman likes \( r \) men, and each man likes \( s \) women. We need to determine the conditions under which there must be at least one man and one woman who like each other.

2. **Apply the Pigeonhole Principle:**
   Consider the men and the women as two sets. Each man likes \( s \) women. By the Pigeonhole Principle, if each of the \( n \) men selects \( s \) women, then there must be at least one woman who is liked by at least \( \lceil \frac{ns}{n} \rceil = s \) men.

3. **Analyze the woman's preferences:**
   Let’s focus on a particular woman who is liked by at least \( s \) men. This woman likes \( r \) men. To ensure that there is at least one man among these \( s \) men who also likes her, we need to consider the number of men who do not like her.

4. **Count the men who do not like her:**
   The total number of men is \( n \). If \( s \) men like her, then \( n - s \) men do not like her.

5. **Ensure overlap in preferences:**
   For the woman to be forced to select a man who likes her, the number of men she likes (\( r \)) must be greater than the number of men who do not like her (\( n - s \)). This ensures that there is at least one man among the \( r \) men she likes who also likes her.

6. **Formulate the inequality:**
   Therefore, we need:
   \[
   r > n - s
   \]
   Rearranging this inequality, we get:
   \[
   r + s > n
   \]

7. **Conclusion:**
   The values of \( r \) and \( s \) that satisfy the condition are those for which:
   \[
   r + s > n
   \]
   Additionally, \( r \) and \( s \) must be within the range \( 1 \leq r, s \leq n \).

The final answer is \( \boxed{ r + s > n } \)