$$Problem 4:$$The sum of $5$ positive numbers equals $2$. Let $S_k$ be the sum of the $k-th$ powers of
these numbers. Determine which of the numbers $2,S_2,S_3,S_4$ can be the greatest among them.
1. **Given Information:**
   - The sum of 5 positive numbers equals 2.
   - Let \( S_k \) be the sum of the \( k \)-th powers of these numbers.
   - We need to determine which of the numbers \( 2, S_2, S_3, S_4 \) can be the greatest among them.

2. **Sum of the First Powers:**
   - Given \( S_1 = x_1 + x_2 + x_3 + x_4 + x_5 = 2 \).

3. **Using Cauchy-Schwarz Inequality for \( S_2 \):**
   - By the Cauchy-Schwarz inequality:
     \[
     (x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2)^2 \leq (x_1 + x_2 + x_3 + x_4 + x_5)(x_1^3 + x_2^3 + x_3^3 + x_4^3 + x_5^3)
     \]
   - Substituting \( S_1 = 2 \):
     \[
     S_2^2 \leq 2 S_3
     \]
   - Therefore, \( S_2 \leq \sqrt{2 S_3} \).

4. **Using Cauchy-Schwarz Inequality for \( S_3 \):**
   - By the Cauchy-Schwarz inequality:
     \[
     (x_1^3 + x_2^3 + x_3^3 + x_4^3 + x_5^3)^2 \leq (x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2)(x_1^4 + x_2^4 + x_3^4 + x_4^4 + x_5^4)
     \]
   - Substituting \( S_2 \) and \( S_4 \):
     \[
     S_3^2 \leq S_2 S_4
     \]
   - Therefore, \( S_3 \leq \sqrt{S_2 S_4} \).

5. **Conclusion:**
   - From the inequalities derived:
     - \( S_2 \leq \sqrt{2 S_3} \)
     - \( S_3 \leq \sqrt{S_2 S_4} \)
   - These inequalities imply that neither \( S_2 \) nor \( S_3 \) can be the greatest among \( 2, S_2, S_3, S_4 \).
   - Therefore, the greatest number can be either \( 2 \) or \( S_4 \).

The final answer is \( \boxed{ 2 } \) or \( S_4 \).