A sequence of $0$s and $1$s

Partial sums. Let $S_k = a_1+\cdots+a_k$ for $k=1,\ldots,n+1$ (with $S_0=0$).

Residues. Consider $S_0,S_1,\ldots,S_{n+1}$: that's $n+2$ values, but only $n$ residues mod $n$. So two are equal: $S_i\equiv S_j\pmod n$ with $i<j$.

Block. Then $a_{i+1}+\cdots+a_j = S_j-S_i\equiv 0\pmod n$.

Conclusion. A consecutive block has sum divisible by $n$.