Show that for all except perfect powers of 2, it's true. 
Let $T_n$ denotes the least natural such that
$$n\mid 1+2+3+\cdots +T_n=\sum_{i=1}^{T_n} i$$
Find all naturals $m$ such that $m\ge T_m$.

I think it’s true for any $m$ except perfect powers of $2.$
For perfect powers of $2.$
Let $m=2^k$. We must have $2^{k+1} \mid T_m(T_m+1)$, hence $2^{k+1} \mid T_m$ or
$2^{k+1} \mid T_m+1$. In both cases, $T_m \geq 2^{k+1}_1 >2^k=m$, therefore all powers of $2$ don't satisfy.
 A: As Adam Rubinson points out in a comment, you also have to show that $n\ge T_n$ when $n$ is not a power of $2$. So given such a positive integer $n$, we want to find $m\le n$ such that $n\mid T_m=\frac12 m(m+1)$, i.e. $2n\mid m(m+1)$.
So write $n$ as $n=a\cdot2^b$ where $a$ is odd. Then we want $a\cdot 2^{b+1}\mid m(m+1)$. We will construct $m\le n$ such that either $a\mid m+1$ and $2^{b+1}\mid m$; or $a\mid m$ and $2^{b+1}\mid m+1$.
Let $c$ be the inverse of $a\bmod 2^{b+1}$, so that $2^{b+1}\mid ac-1$.

*

*If $c\le 2^b$, put $m=ac-1$. Then $a\mid ac=m+1$ and $2^{b+1}\mid m$; also, $m\le a\cdot 2^b-1<n$.


*If $c>2^b$, put $m=2n-ac$. Then $a\mid n$, so $a\mid m$; and $2^{b+1}$ divides both $2n$ and $ac-1$, so $2^{b+1}\mid 2n-(ac-1)=m+1$. Also, $m=2n-ac<2n-a\cdot 2^b=n$.
For instance, suppose $n=112=7\cdot 2^4$. Then $c=7^{-1} \bmod 32=23$. This is greater than $2^b=16$, so we put $m=2n-ac=224-7\cdot 23=63$. And you can check that $224\mid 63\cdot 64$.
I have deliberately left a logical gap here, to give you something to think about: where does this process go wrong when $n$ is a power of $2$? All the steps look valid, even if $a=1$. So where is the gap?
