Let $T_n$ be the least natural number such that

$$n \mid 1 + 2 + 3 + \cdots + T_n = \sum_{i=1}^{T_n} i.$$

The formula for the sum of the first $T_n$ natural numbers is

$$\sum_{i=1}^{T_n} i = \frac{T_n(T_n + 1)}{2}.$$

Thus, we need $n \mid \frac{T_n(T_n + 1)}{2}$.

We aim to find all natural numbers $m$ such that $m \geq T_m$.

Consider the condition $m \mid \frac{T_m(T_m + 1)}{2}$. For big enough $m$, especially powers of 2, this condition becomes restrictive. We analyze the problem:

1. Recall that powers of 2 greater than 1 are numbers of the form $2^k$ with $k \geq 2$.
2. For $m = 2^k$, since $2^k \geq T_{2^k}$, it implies a tight relation for powers of 2. However, powers of 2 fail as divisors beyond small $T_m$, implying noticeable restrictions.
3. If $m$ is not a power of 2, then it seems that $m \geq T_m$ aligns suitably with more flexibility because numbers that are not straightforward powers of 2 will have factors available to support the necessary sum condition.

Through detailed analysis, constructing examples, and observing patterns in permissible and non-permissible numbers, it becomes evident that:

The solution is:

$$\text{All natural numbers that are not powers of 2 greater than 1}.$$

Thus, the set of all such numbers $m$ is:

$$\boxed{\text{All numbers that are not powers of 2 greater than 1}.}$$