Let $n\ge3$ be a positive integer. Alex and Lizzie play a game. Alex chooses $n$ positive integers (not necessarily distinct), writes them on a blackboard, and does nothing further. Then, Lizzie is allowed to pick some of the numbers―but not all of them―and replace them each by their average. For example, if $n=7$ and the numbers Alex writes on the blackboard to start are $1,2,4,5,9,4,11$, then on her first turn Lizzie could pick $1,4,9$, erase them, and replace them each with the number $\tfrac{1+4+9}{3}$, leaving on the blackboard the numbers $\tfrac{14}{3},2,\tfrac{14}{3},5,\tfrac{14}{3},4,11$. Lizzie can repeat this process of selecting and averaging some numbers as often as she wants. Lizzie wins the game if eventually all of the numbers written on the blackboard are equal. Find all positive integers $n\ge3$ such that no matter what numbers Alex picks, Lizzie can win the game.
1. **Claim:** Composite \( n \) work.

   **Proof:** 
   - Let \( n \) be a composite number. This means \( n \) has a proper divisor \( k \) such that \( 1 < k < n \).
   - Partition the \( n \) integers into \( k \) sets \( S_1, S_2, \dots, S_k \), each containing \( \frac{n}{k} \) elements.
   - Average the elements in each set \( S_i \). Let \( a_i \) be the average of the elements in \( S_i \).
   - Replace each element in \( S_i \) with \( a_i \). Now, the board contains \( \frac{n}{k} \) copies of each \( a_i \).
   - Rearrange these into \( \frac{n}{k} \) identical sets of the form \( \{a_1, a_2, \dots, a_k\} \).
   - Average each set \( \{a_1, a_2, \dots, a_k\} \) to get \( \frac{a_1 + a_2 + \dots + a_k}{k} \).
   - Repeat this process until all numbers on the board are equal.

   This shows that for composite \( n \), Lizzie can always win the game. \(\blacksquare\)

2. **Claim:** Prime \( n \) fail.

   **Proof:**
   - Suppose \( n = p \) is a prime number.
   - Alex writes \( p-1 \) ones and one two on the board. The initial numbers are \( 1, 1, \dots, 1, 2 \).
   - The average of these numbers is \( \frac{(p-1) \cdot 1 + 2}{p} = \frac{p+1}{p} \).
   - Lizzie aims to convert all numbers into \( \frac{p+1}{p} \).

   To show that Lizzie cannot achieve this:
   - Consider the denominators of the numbers on the board. Initially, all numbers are integers, so their denominators are 1.
   - When Lizzie averages \( k \) numbers with denominators \( d_1, d_2, \dots, d_k \), the denominator of the resulting average is a factor of \( k \cdot \text{lcm}(d_1, d_2, \dots, d_k) \).
   - Since \( k < p \) and \( p \) is prime, \( p \) does not divide \( k \).
   - By induction, assume that at some moment, none of the numbers have a denominator divisible by \( p \).
   - Averaging \( k \) numbers with denominators \( d_1, d_2, \dots, d_k \), the new denominator is a factor of \( k \cdot \text{lcm}(d_1, d_2, \dots, d_k) \), which is not divisible by \( p \).
   - Therefore, at no point can a number on the board have a denominator divisible by \( p \).

   Since \( \frac{p+1}{p} \) has a denominator \( p \), Lizzie cannot achieve this, and Alex wins the game when \( n \) is prime. \(\blacksquare\)

The final answer is \(\boxed{\text{composite } n}\)