For integer $n\ge 4$, find the minimal integer $f(n)$, such that for any positive integer $m$, in any subset with $f(n)$ elements of the set ${m, m+1, \ldots, m+n+1}$ there are at least $3$ relatively prime elements.
1. **Initial Consideration**:
   For \( n \geq 4 \), we need to find the minimal integer \( f(n) \) such that any subset with \( f(n) \) elements of the set \(\{m, m+1, \ldots, m+n+1\}\) contains at least 3 relatively prime elements.

2. **Upper Bound**:
   Consider the set \( M = \{m, m+1, m+2, \ldots, m+n-1\} \).
   - If \( 2 \mid m \), then \( m+1, m+2, m+3 \) are mutually prime.
   - If \( 2 \nmid m \), then \( m, m+1, m+2 \) are mutually prime.
   Therefore, in every \( n \)-element subset of \( M \), there are at least 3 mutually prime elements. Hence, \( f(n) \leq n \).

3. **Specific Cases**:
   - **Case \( n = 4 \)**:
     Consider \(\{m, m+1, m+2, m+3\}\).
     - If \( m \) is odd, then \( m, m+1, m+2 \) are mutually prime.
     - If \( m \) is even, then \( m+1, m+2, m+3 \) are mutually prime.
     Thus, \( f(4) \leq 4 \). However, choosing a 3-element subset consisting of two evens and one odd shows these three numbers are not mutually prime. Hence, \( f(4) = 4 \).

   - **Case \( n = 5 \)**:
     Consider \(\{m, m+1, m+2, m+3, m+4\}\).
     - If \( m \) is even, then \( m, m+2, m+4 \) are all even.
     Any 3 numbers from the 4-element subset \(\{m, m+1, m+2, m+4\}\) are not mutually prime, so \( f(5) > 4 \). But from the 5-element set, we can find 3 numbers which are mutually prime. Thus, \( f(5) = 5 \).

   - **Case \( n = 6 \)**:
     Consider \(\{m, m+1, m+2, m+3, m+4, m+5\}\).
     - The set contains 3 odd and 3 even numbers.
     If we choose a 4-element subset consisting of 3 evens and 1 odd, then among the subset there are no 3 numbers which are mutually prime. Thus, \( f(6) > 4 \). Consider the 5-element subset:
     - If among these 5 elements there are 3 odds, then these 3 numbers must be mutually prime.
     - If these 5 elements are 3 evens and 2 odds, then among these 3 evens, there is at most one number divisible by 3 and at most one number divisible by 5. Hence, among these 3 evens, there is one number which is neither divisible by 3 nor 5. The even number and the other 2 odds are mutually prime. Thus, \( f(6) = 5 \).

4. **Generalization**:
   - **Inductive Step**:
     Note that \(\{m, m+1, \ldots, m+n\} = \{m, m+1, \ldots, m+n-1\} \cup \{m+n\}\) implies \( f(n+1) \leq f(n) + 1 \).

   - **Lower Bound**:
     For \( n > 6 \), let \( X_n = \{x \mid x \leq n+1 \text{ and } 2 \mid x \text{ or } 3 \mid x\} \). Then \( X_n \) is a subset of \(\{2, 3, \ldots, n+1\}\), and any 3 elements in \( X_n \) are not mutually prime. Thus, \( f(n) \geq |X_n| + 1 \). By the inclusion-exclusion principle:
     \[
     |X_n| = \left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+1}{3} \right\rfloor - \left\lfloor \frac{n+1}{6} \right\rfloor
     \]
     Therefore,
     \[
     f(n) \geq \left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+1}{3} \right\rfloor - \left\lfloor \frac{n+1}{6} \right\rfloor + 1
     \]

5. **Final Calculation**:
   Let \( n = 6k + r \), \( k \geq 1 \), \( r = 0, 1, 2, 3, 4, 5 \). Then:
   \[
   \left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+1}{3} \right\rfloor - \left\lfloor \frac{n+1}{6} \right\rfloor + 1 = 4k + \left\lfloor \frac{r+1}{2} \right\rfloor + \left\lfloor \frac{r+1}{3} \right\rfloor - \left\lfloor \frac{r+1}{6} \right\rfloor + 1
   \]
   And:
   \[
   \left\lfloor \frac{r+1}{2} \right\rfloor + \left\lfloor \frac{r+1}{3} \right\rfloor - \left\lfloor \frac{r+1}{6} \right\rfloor + 1 = 
   \begin{cases} 
   r, & \text{if } r = 0, 1, 2, 3 \\
   r - 1, & \text{if } r = 4, 5 
   \end{cases}
   \]

   - **For \( r = 0, 1, 2, 3 \)**:
     We can divide \( n = 6k + r \) numbers into \( k \) groups \(\{m, m+1, \ldots, m+5\}, \{m+6, m+7, \ldots, m+11\}, \ldots, \{m+6(k-1), m+6k-5, \ldots, m+6k-1\}\), and we are left with \( r \) numbers \( m+6k, \ldots, m+6k+r-1 \). Among \( 4k + r + 1 \) numbers, there are at least \( 4k + 1 \) numbers which are contained in the above \( k \) groups. Thus, there is at least one group which contains 5 numbers. Since \( f(6) = 5 \), there are 3 numbers which are mutually prime.

   - **For \( r = 4, 5 \)**:
     We can prove similarly that there are 3 numbers which are mutually prime.

   Hence,
   \[
   f(n) = \left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+1}{3} \right\rfloor - \left\lfloor \frac{n+1}{6} \right\rfloor + 1
   \]

The final answer is \( \boxed{ \left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+1}{3} \right\rfloor - \left\lfloor \frac{n+1}{6} \right\rfloor + 1 } \)