Number $125$ is written as the sum of several pairwise distinct and relatively prime numbers, greater than $1$. What is the maximal possible number of terms in this sum?
1. **Identify the problem constraints:**
   We need to express the number \(125\) as the sum of several pairwise distinct and relatively prime numbers, each greater than \(1\). We aim to find the maximum number of such terms.

2. **Construct a potential solution:**
   Consider the sum \(3 \times 2 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 125\). This sum uses 8 terms. We need to verify if 8 is indeed the maximum number of terms.

3. **Check the sum of the first 9 smallest odd primes:**
   Since \(2\) is the only even prime and including it would make the sum even, which contradicts the odd sum \(125\), we only consider odd primes. The first 9 odd primes are \(3, 5, 7, 11, 13, 17, 19, 23, 29\).

   Calculate their sum:
   \[
   3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 127
   \]
   Since \(127 > 125\), it is impossible to use 9 terms.

4. **Verify the sum of the first 8 smallest odd primes:**
   The first 8 odd primes are \(3, 5, 7, 11, 13, 17, 19, 23\).

   Calculate their sum:
   \[
   3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 98
   \]
   Since \(98 < 125\), we need to find another combination of 8 terms that sum to \(125\).

5. **Construct a valid combination of 8 terms:**
   Consider the combination \(3 \times 2 + 7 + 11 + 13 + 17 + 19 + 23 + 29\):
   \[
   6 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 125
   \]
   This combination sums to \(125\) and uses 8 terms.

6. **Prove that 8 is the maximum number of terms:**
   Since the sum of the first 9 smallest odd primes exceeds \(125\) and the sum of the first 8 smallest odd primes is less than \(125\), we conclude that 8 is the maximum number of terms.

\[
\boxed{8}
\]