This one might be right, but I might be wrong.

Mathematical induction

When n=2, obvious.

Assume the statement is true for n.

For n+1, write M as {b1,b2,…b(n)}, where b1
Let a(n+1) be the max of a(i).

case 1:
If b(n-1)>a1+a2+ … + a(n).
By induction, a solution exists for {a1,…, a(n)} and {b1,…, b(n-1)}. Adding a(n+1) and b(n) won’t cause any conflict.

Case 2:
If b(n-1) By induction, a solution exists for {a1,…, a(n)} and {b1,…, b(n-1)}. If there exists k<=n such that b(n)=a1+…+a(k), insert a(n+1) in front of a(k).

Case 3:
If b(n-1)=a1+a2+…a(n).
If b(n-2)=a1+…+a(n-1), then b(n-1)-b(n-2)=a(n). A solution (based on induction) for {a1,a2,…a(n-1), a(n+1)} and {b1,b2,…,b(n-2),b(n)} can be used to form a solution for {a1,a2,…a(n), a(n+1)} and {b1,b2,…,b(n-1),b(n)}. So we assume b(n-2) is not equal to a1+…+a(n-1). Therefore, a solution exists. This solution plus a(n+1) is bigger b(n-1). If it happens to be b(n), letting a different number be a(n) will avoid it.