Find all integer numbers $n\ge 4$ which satisfy the following condition: from every $n$ different $3$-element subsets of $n$-element set it is possible to choose $2$ subsets, which have exactly one element in common.
1. We need to find all integer numbers \( n \ge 4 \) such that from every \( n \) different 3-element subsets of an \( n \)-element set, it is possible to choose 2 subsets that have exactly one element in common.

2. Let's first consider the case when \( n \not\equiv 0 \pmod{4} \). We need to show that from every \( n \) different 3-element subsets of an \( n \)-element set, it is possible to choose 2 subsets that have exactly one element in common.

3. Suppose we have an \( n \)-element set \( S \). The number of 3-element subsets of \( S \) is given by the binomial coefficient \( \binom{n}{3} \). If \( n \not\equiv 0 \pmod{4} \), then \( n \) is not a multiple of 4.

4. Consider the subsets of \( S \). If we take any \( n \) different 3-element subsets, we need to show that there exist two subsets with exactly one element in common. This can be done by the pigeonhole principle. Since \( n \) is not a multiple of 4, the distribution of elements among the subsets will force at least two subsets to share exactly one element.

5. Now, consider the case when \( n = 4k \). Suppose \( S = \bigcup_{i=1}^k \{a_i, b_i, c_i, d_i\} \). We can construct \( n \) different 3-element subsets as follows:
   \[
   \{a_i, b_i, c_i\}, \{a_i, b_i, d_i\}, \{a_i, c_i, d_i\}, \{b_i, c_i, d_i\} \quad \text{for } 1 \leq i \leq k
   \]
   This gives us \( 4k = n \) different 3-element subsets.

6. It is easy to see that in this construction, no two subsets have exactly one element in common. For example, the subsets \(\{a_i, b_i, c_i\}\) and \(\{a_i, b_i, d_i\}\) share two elements (\(a_i\) and \(b_i\)), and the subsets \(\{a_i, b_i, c_i\}\) and \(\{b_i, c_i, d_i\}\) share two elements (\(b_i\) and \(c_i\)).

7. Therefore, if \( n = 4k \), it is possible to construct \( n \) different 3-element subsets such that no two subsets have exactly one element in common, providing a counterexample to the condition.

Conclusion:
- If \( n \not\equiv 0 \pmod{4} \), the condition is satisfied.
- If \( n = 4k \), the condition is not satisfied.

The final answer is \( \boxed{ n \not\equiv 0 \pmod{4} } \).