Suppose that $125=a_1+a_2+...+a_n$ where $a_i$ are pairwise coprime positive integers. Clearly there are at most one even number. And at least $n-1$ odd numbers. But the sum of the first 12 odd numbers is $144>125$. Thus there are at most 11 odd number and one even number. Hence $n\le 12$. In the case $n=12$ we must have an even number and 11 odd numbers. The sum of the odd numbers from 1 to 21 with the smallest even number 2 is 123 And if we change any of the odd numbers the sum will increase by 2. But to make a list of 11 odd pairwise coprime numbers we need to change more than one number. Hence we can't find such numbers. We can consider the case $n=11$ in the same way (by number of changes which we need). In the case $n=10$ the numbers are : $$125=1+2+3+5+7+11+13+17+29+37$$thus the maximum is 10.

Euleralfeel wrote: In the case $n=10$ the numbers are : $$125=1+2+3+5+7+11+13+17+29+37$$thus the maximum is 10. $1$ cannot be a term of the sum. Leicich wrote: a sum of some pairwise coprime integers larger than $1$.

Oh !!! I am sorry I think that It can be

Since the numbers are coprime , then each number has at least one prime factor and if there are 10 coprime numbers then their sum must be greatest or equal to the sum of the first 10 primes : $$2+3+5+7+11+13+17+23+29=129$$The case $n=9$ we need 9 odd numbers to make an odd sum but the sum of the first 9 odd primes is greatest than 125 so we can't find such integers. The case $n=8$ the numbers are : $$8+3+5+7+11+13+17+61=125$$