Answer: $\frac{n(n-1)}{2}$.

Each set of segments that satisfies the condition of the problem is called good. Consider a set consisting of all segments of the form $[a, b]$, where $a, b \in \mathbb{N}$ and $a < b \le n$. This collection contains $n(n-1)/2$ segments and it is good since even the union of all segments of the collection does not cover the point $0$. Therefore the maximum possible value of the number $m$ is not less than $n(n-1)/2$.

Let us prove the following
Claim: For any good collection $A$ there exists a good collection $B$ containing the same number of segments and not containing any segment of the form $[0, k]$, $k = 1, \ldots, n$.

Proof of the claim: If the set $A$ does not contain any segment of the form $[0, k]$, then put $B = A$. Otherwise among all segments of $A$ of the form $[0, k]$, consider the segment $[0, s]$ of minimum length. Let us exclude the segment $[0, s]$ from the set and instead add the segment $[s, n]$, which obviously was not contained in $A$. The resulting set will also be good. Indeed if it's possible to select several segments of the total length $n$ from the new set, the union of which would coincide with the entire segment $[0, n]$, then among them there must be a segment $[s, n]$, but then several non-intersecting segments would cover the segment $[0, s]$ and among them there would be a segment of the form $[0, t]$ where $t < s$, which is impossible due to the minimality of the number $s$. Repeating the described procedure several times, we obtain the required set $B$. The assertion is proved.

From the proved statement, in particular, it follows that the number of segments in an arbitrary good set is not more than $n(n-1)/2$, and hence the maximum possible value of the number $m$ is exactly equal to $n(n-1)/2$.