A family of sets $ A_1, A_2, \ldots ,A_n$ has the following properties:

[b](i)[/b] Each $ A_i$ contains 30 elements.
[b](ii)[/b] $ A_i \cap A_j$ contains exactly one element for all $ i, j, 1 \leq i < j \leq n.$

Determine the largest possible $ n$ if the intersection of all these sets is empty.
1. **Understanding the Problem:**
   We are given a family of sets \( A_1, A_2, \ldots, A_n \) with the following properties:
   - Each set \( A_i \) contains exactly 30 elements.
   - The intersection of any two distinct sets \( A_i \) and \( A_j \) contains exactly one element.
   - The intersection of all these sets is empty.

2. **Analyzing the Properties:**
   - Since each \( A_i \) contains 30 elements, and each pair \( A_i \cap A_j \) contains exactly one element, we need to determine the largest possible \( n \) such that the intersection of all sets is empty.

3. **Counting Elements:**
   - Let \( S \) be the union of all sets \( A_i \). We need to count the number of elements in \( S \).
   - Each element in \( S \) must appear in at least one of the sets \( A_i \). Since the intersection of all sets is empty, no element is common to all sets.

4. **Using Combinatorial Argument:**
   - Consider the total number of pairwise intersections. There are \( \binom{n}{2} \) such intersections, and each intersection contains exactly one element.
   - Therefore, there are \( \binom{n}{2} \) distinct elements that are counted in the intersections.

5. **Counting Elements in Each Set:**
   - Each set \( A_i \) contains 30 elements. Since each element in \( A_i \) is shared with exactly \( n-1 \) other sets, each element in \( A_i \) is counted \( n-1 \) times in the union of all sets.

6. **Total Number of Elements:**
   - The total number of elements in \( S \) can be calculated by considering the number of elements in each set minus the overcounting due to intersections.
   - Let \( |S| \) be the number of distinct elements in \( S \). We have:
     \[
     |S| = 30n - \binom{n}{2}
     \]
   - Simplifying the binomial coefficient:
     \[
     \binom{n}{2} = \frac{n(n-1)}{2}
     \]
   - Therefore:
     \[
     |S| = 30n - \frac{n(n-1)}{2}
     \]

7. **Ensuring Non-Negative Elements:**
   - Since \( |S| \) must be non-negative, we set up the inequality:
     \[
     30n - \frac{n(n-1)}{2} \geq 0
     \]
   - Multiplying through by 2 to clear the fraction:
     \[
     60n - n^2 + n \geq 0
     \]
   - Rearranging terms:
     \[
     n^2 - 61n \leq 0
     \]
   - Factoring the quadratic expression:
     \[
     n(n - 61) \leq 0
     \]

8. **Solving the Inequality:**
   - The inequality \( n(n - 61) \leq 0 \) holds when \( 0 \leq n \leq 61 \).

9. **Conclusion:**
   - The largest possible \( n \) that satisfies the given conditions is \( n = 61 \).

The final answer is \( \boxed{61} \)