Let $A=\{1,2,...,100\}$ and $f(k), k\in N$ be the size of the largest subset of $A$ such that no two elements differ by $k$. How many solutions are there to $f(k)=50$?
1. **Define the set and function:**
   Let \( A = \{1, 2, \ldots, 100\} \) and \( f(k) \) be the size of the largest subset of \( A \) such that no two elements differ by \( k \).

2. **Property \( P1 \):**
   In a set containing \( k \) consecutive natural numbers, the difference between any two numbers is less than \( k \).

3. **Property \( P2 \):**
   For \( k \leq 50 \), in a set \( S = \{a, a+1, a+2, \ldots, a+2k-1\} \) containing \( 2k \) consecutive natural numbers, there exist at most \( k \) numbers such that the difference between any two numbers is different from \( k \). Indeed, at most one of the numbers \( b, b+k \in S \) can appear in \( B \), where \( a \leq b \leq a+k-1 \). Thus, \( |S \cap B| \leq k \).

4. **Property \( P3 \):**
   For \( k \geq 51 \), using \( P1 \), we can choose \( \{1, 2, \ldots, k\} \subset B \), hence \( f(k) \geq |B| \geq k \geq 51 \). Therefore, \( f(k) = 50 \) can have solutions only for \( k \leq 50 \).

5. **Case 1: \( k \mid 50 \):**
   - There exists \( d \in \mathbb{N} \) such that \( kd = 50 \).
   - Denote \( A_m = \{2mk+1, 2mk+2, \ldots, (2m+2)k\} \) for \( 0 \leq m \leq d-1 \). We observe that \( A = \cup_{m=0}^{d-1} A_m \). Using property \( P2 \), we get \( |A_m \cap B| \leq k \), which implies \( |A \cap B| = |B| \leq kd = 50 \).
   - Construct \( B = \cup_{m=0}^{d-1} B_m = \cup_{m=0}^{d-1} \{2mk+1, 2mk+2, \ldots, (2m+1)k\} \).
   - In each subset \( \{2mk+1, 2mk+2, \ldots, (2m+1)k\} \), the difference between any two numbers is less than \( k \). For \( p, q \) elements in different subsets \( B_m \), we have \( |p-q| > k \). Thus, \( |B| = kd = 50 \).
   - Therefore, the set \( B \) constructed above has the maximum size.
   - Hence, all \( k \in \mathbb{N} \) such that \( k \mid 50 \) are solutions of the equation \( f(k) = 50 \).

6. **Case 2: \( k < 50 \) and \( k \nmid 50 \):**
   - There exist \( d, r \in \mathbb{N} \) such that \( 1 \leq r \leq k-1 \) and \( 50 = kd + r \). This implies \( 100 = 2kd + 2r \).
   - For \( 0 \leq m \leq d-1 \), denote \( B_m = \{2mk+1, 2mk+2, \ldots, (2m+1)k\} \).
   - Let \( B_d = \{2kd+1, 2kd+2, \ldots, \min(2kd+k, 100)\} \).
   - The set \( B = \cup_{m=0}^{d} B_m \) has \( |B| = \min(kd+k, kd+100-2kd) \). Since \( kd+k > kd+r = 50 \) and \( kd+100-2kd = kd+2r > kd+r = 50 \), we have \( f(k) \geq |B| > 50 \).
   - Therefore, \( k \) is not a solution of \( f(k) = 50 \).

7. **Conclusion:**
   The solutions of \( f(k) = 50 \) are the numbers \( k \) such that \( k \mid 50 \). Since \( 50 = 2 \cdot 5^2 \), \( 50 \) has \( (1+1)(2+1) = 6 \) natural divisors. Thus, \( f(k) = 50 \) has 6 solutions: \( k \in \{1, 2, 5, 10, 25, 50\} \).

The final answer is \( \boxed{ \{1, 2, 5, 10, 25, 50\} } \)