Let $S$ be a subset of $\{0,1,2,\ldots,98 \}$ with exactly $m\geq 3$ (distinct) elements, such that for any  $x,y\in S$  there exists  $z\in S$  satisfying $x+y \equiv 2z \pmod{99}$. Determine all possible values of $m$.
1. **Given Condition and Initial Setup:**
   We are given a subset \( S \) of \(\{0,1,2,\ldots,98\}\) with exactly \( m \geq 3 \) distinct elements. The condition is that for any \( x, y \in S \), there exists \( z \in S \) such that \( x + y \equiv 2z \pmod{99} \).

2. **Case 1: \( m \mid 99 \):**
   - Let \( 99 = m \cdot d \). Consider the set \( S = \{0, d, 2d, \ldots, (m-1)d\} \).
   - For any \( x = ud \) and \( y = vd \) in \( S \), we have \( x + y = (u+v)d \).
   - We need \( z \) such that \( 2z \equiv x + y \pmod{99} \). Let \( z = wd \). Then \( 2wd \equiv (u+v)d \pmod{99} \).
   - This simplifies to \( 2w \equiv u + v \pmod{m} \), which always has a solution \( w \) in \(\{0, 1, 2, \ldots, m-1\}\) because \( m \) is odd.
   - Therefore, if \( m \mid 99 \), the condition is satisfied.

3. **Case 2: \( m \nmid 99 \):**
   - Let \( x_1, x_2, \ldots, x_m \) be the elements of \( S \).
   - Consider the sums \( x_m + x_1, x_m + x_2, \ldots, x_m + x_{m-1} \) and let \( x_m + x_i \equiv 2y_i \pmod{99} \) for \( i = 1, 2, \ldots, m-1 \).
   - Since \( x_i \)'s are distinct modulo 99, the \( y_i \)'s are also distinct modulo 99.
   - None of the \( y_i \)'s can be equivalent to \( x_m \) modulo 99 because otherwise, we would have two equivalent terms.
   - Thus, \( (y_1, y_2, \ldots, y_{m-1}) \) is a permutation of \( (x_1, x_2, \ldots, x_{m-1}) \).
   - Adding these equations, we get \( (m-1)x_m \equiv x_1 + x_2 + \ldots + x_{m-1} \pmod{99} \).
   - This implies \( mx_m \equiv x_1 + x_2 + \ldots + x_m \pmod{99} \).
   - Repeating this for all \( x_i \)'s, we get \( mx_1 \equiv mx_2 \equiv \ldots \equiv mx_m \pmod{99} \).
   - Let \( \gcd(m, 99) = d \) and \( m = dm_1 \), \( 99 = dk \). Then \( x_1 \equiv x_2 \equiv \ldots \equiv x_m \pmod{k} \).
   - We have \( m \) numbers equivalent to each other modulo \( k \). Since there are \( d \) numbers modulo \( kd \) equivalent to each other modulo \( k \), we must have \( m \leq d \).
   - This implies \( dm_1 \leq d \), so \( m_1 = 1 \). Hence, \( m = d \mid 99 \).

4. **Conclusion:**
   - The possible values of \( m \) are the divisors of 99 that are greater than or equal to 3.
   - The divisors of 99 are 1, 3, 9, 11, 33, and 99. Excluding 1 and including those greater than or equal to 3, we get \( m = 3, 9, 11, 33, 99 \).

The final answer is \( \boxed{ m = 3, 9, 11, 33, 99 } \).