A multiple of $n$ from $0$s and $1$s

Sequence. Consider $a_k = \underbrace{1\cdots1}_{k}$ for $k=1,\ldots,n+1$.

Pigeonhole. $n+1$ numbers and $n$ residues mod $n$: two have the same residue, say $a_i\equiv a_j\pmod n$ with $i<j$.

Difference. $a_j-a_i = \underbrace{1\cdots1}_{j-i}\underbrace{0\cdots0}_{i}$, which is a multiple of $n$ and uses only $0$s and $1$s.

Conclusion. A multiple with the required form exists.