Suppose not. $\implies \exists m\in \mathbb{N}$ s.t. $m\geq2$ and $\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{m}=S\in\mathbb{N}$.
Pick $k\in\mathbb{N}$ s.t. $2^k\leq m$ but $2^{k+1}>m$: so $2^k$ is the greatest power less than or equal to $m$.
And if $m=2^lm'$ with $m'$ odd, we have $k\geq l$
$\implies 2^kS=\frac{2^k}{1}+\frac{2^k}{2}+\cdots+\frac{2^k}{m}$
$\implies 2^kS = \frac{2^k}{1}+\frac{2^{k-1}}{1}+\frac{2^k}{3}+\cdots+1+\cdots+\frac{2^{k-l}}{m'}$ (if $m=2^k$, then $\frac{2^k}{m}$ is the $1$, and no $1$ in the middle)
Let $L=$ lcm of the denominators of RHS: $L$ is an odd number.
$\implies 2^kSL=2^kL+2^{k-1}L+2^k\frac{L}{3}+\cdots+L+\cdots+2^{k-l}\frac{L}{m'}$ (note that each fraction of the form $\frac{L}{x}$ is an integer)
$\implies 2^kSL=2\left(2^{k-1}L+2^{k-2}L+2^{k-1}\frac{L}{3}+\cdots+2^{k-l-1}\frac{L}{m'}\right)+L$
(if $l=k$, then $2^{k-l-1}\frac{L}{m'}$ is not inside the parenthesis since $m=2^lm'=2^km'=2^k\implies 2^{k-l}\frac{L}{m'}=L$)
We could factor $2$ out from the RHS since before multiplying $2^k$, originally each denominator contained at most $k-1$ power of $2$ except $2^k$ itself. And the $L$ term on the RHS was that $1/2^k$.
Since $k\geq1$, LHS is an even number while RHS is an odd number. Which is a contradiction.
From above, did I miss anything or is there something unclear/wrong?