4. Given that $S$ is a set composed of $n(n \geqslant 3)$ positive numbers. If there exist three distinct elements in $S$ that can form the three sides of a triangle, then $S$ is called a "triangle number set". Consider the set of consecutive positive integers $\{4,5, \cdots, m\}$, all of whose 10-element subsets are triangle number sets. Then the maximum possible value of $m$ is
4. 253

Analysis: When $m=254$, the ten-element set $\{4,5,9,14,23,37,60,97,157,254\}$ does not contain three numbers that can form the three sides of a triangle, so $m \leq 253$. We now prove that $m=253$ is feasible.

The following proof uses proof by contradiction: Assume that when $m=253$, there exists a ten-element set $S$ such that no three elements in $S$ can form the three sides of a triangle.

Without loss of generality, let $S=\left\{a_{1}, a_{2}, \ldots, a_{10}\right\}, 4 \leq a_{1} < a_{2} < \ldots < a_{10} \leq 253$. Then, $a_{10}+a_{9}>2 a_{9}+a_{8}>2 a_{8}+a_{7}>3 a_{7}+2 a_{6}>\ldots>34 a_{2}+21 a_{1} \geq 34 \times 5+21 \times 4=254$, which is a contradiction.

Therefore, when $m=253$, all ten-element sets $S$ satisfy: there exist three elements in $S$ that can form the three sides of a triangle.