For $M$ to be $n$-balanced, the number of its non-empty subsets must be a multiple of $n$, so $s=|M|$ must satisfy $n\mid 2^s-1$. Since the sequence $2^m \pmod{n}$ is for odd $n$ (in both directions) periodic and $2^0=1$, such $0<s\leq n$ exists. Take smallest such $0<s\leq n$, let us set $M'=\{2^0,2^1,...,2^{s-1}\}$. Since every positive integer has a unique binary representation, the sums of subsets of $M'$ are exactly the numbers $1,2,3,...,2^s-1$, each appearing exactly once. Since by construction $n\mid 2^s-1$, we can construct $M$ by replacing each $m'\in M'$ by its residue mod $n$. By minimality of $s$, these residues are pairwise distinct and also such $M$ is $n$-balanced. We are done.