Let $n$ be an integer greater than $1$ and let $X$ be an $n$-element set. A non-empty collection of subsets $A_1, ..., A_k$ of $X$ is tight if the union $A_1 \cup \cdots  \cup A_k$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_i$s. Find the largest cardinality of a collection of proper non-empty subsets of $X$, no non-empty subcollection of which is tight.

[i]Note[/i]. A subset $A$ of $X$ is proper if $A\neq X$. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection.
1. **Define the problem and initial setup:**
   Let \( n \) be an integer greater than 1 and let \( X \) be an \( n \)-element set. We need to find the largest cardinality of a collection of proper non-empty subsets of \( X \), no non-empty subcollection of which is tight. A collection of subsets \( A_1, A_2, \ldots, A_k \) is tight if:
   - The union \( A_1 \cup A_2 \cup \cdots \cup A_k \) is a proper subset of \( X \).
   - No element of \( X \) lies in exactly one of the \( A_i \)'s.

2. **Claim and construction:**
   We claim that the largest cardinality of such a collection is \( 2n - 2 \). To construct this, consider the set \( X = \{1, 2, \ldots, n\} \). We can take:
   - All \( n \) subsets of \( X \) with cardinality \( n-1 \).
   - \( n-2 \) single-element subsets.

   This gives us \( n + (n-2) = 2n - 2 \) subsets. 

3. **Verification of the construction:**
   - If the subcollection contains two sets of cardinality \( n-1 \), their union is \( X \), so it is not tight.
   - If the subcollection does not contain two sets of cardinality \( n-1 \), there will always be an element that lies in only the \( n-1 \) element subset chosen, since the described collection only has \( n-2 \) one-element sets.

4. **Inductive argument:**
   - **Base case:** For \( n = 2 \), the set \( X = \{1, 2\} \). The subsets are \(\{1\}\) and \(\{2\}\). The collection \(\{\{1\}, \{2\}\}\) has cardinality 2, which is \( 2n - 2 = 2 \times 2 - 2 = 2 \).
   - **Inductive step:** Assume the statement is true for \( n \). We need to show it holds for \( n+1 \).

5. **Splitting the collection:**
   - Consider a collection \( \mathcal{C} \) of sets that satisfies the problem conditions. For any two elements \( a, b \in X \), there is some set in \( \mathcal{C} \) containing only one of them. If not, we can combine \( a \) and \( b \) into one element and reduce \( |X| \), since we only care about the number of sets in \( \mathcal{C} \) and this reduction does not affect that.
   - Consider an element, say \( 1 \), that appears in the least number of sets in \( \mathcal{C} \). Split the collection \( \mathcal{C} \) into two subcollections, \( \mathcal{A} \) and \( \mathcal{B} \), where \( \mathcal{A} \) consists of sets containing \( 1 \) and \( \mathcal{B} \) consists of sets not containing \( 1 \).

6. **Analyzing subcollections:**
   - There must be some element in \( X \setminus \{1\} \) that appears in only one set in \( \mathcal{B} \), or else \( \mathcal{B} \) would be tight. Suppose this element is \( 2 \) and the unique set it appears in is \( b \in \mathcal{B} \).
   - In subcollection \( \mathcal{A} \), each set contains \( 1 \). There can be at most one set in \( \mathcal{A} \) not containing \( 2 \), or else the number of sets containing \( 1 \) exceeds the number of sets containing \( 2 \), contradicting our assumption of minimality. Let this set, if it exists, be \( a \in \mathcal{A} \).

7. **Combining elements:**
   - Set aside \( a \) and \( b \). Among the remaining sets in \( \mathcal{C} \), all sets either contain both \( 1 \) and \( 2 \) or none of \( 1 \) or \( 2 \). In this case, we can combine \( 1 \) and \( 2 \) into a single element, reducing \( |X| \) by one and deleting at most 2 sets in \( \mathcal{C} \).
   - Repeat this operation until \( |X| \) reduces to 2. Therefore, \( |\mathcal{C}| \leq 2(|X|-2) + 2 = 2|X| - 2 = 2n - 2 \).

Thus, the largest cardinality of a collection of proper non-empty subsets of \( X \), no non-empty subcollection of which is tight, is \( 2n - 2 \).

The final answer is \( \boxed{ 2n - 2 } \).