Let $T_{\mathrm{n}}$ be the number we are looking for, then $T_{\mathrm{n}}+{ }_{1}$ is the corresponding number for the set $S \cup\{n+1\}$. Consider $\Gamma_{n}+1-T_{n}$, it represents the number of triangles with one side length $n+1$ and the other two sides being different numbers from $S$. For each such triangle, the sum of the lengths of the two sides not equal to $n+1$ is one of the following numbers:
$$2 n-1,2 n-2, \cdots, n+3, n+2$$
(there are $n-2$ such numbers).
The number of triangles with these sums are respectively
$$\begin{array}{l} 1,1,2,2,3,3, \cdots \text {, up to } n-2 \text { terms } \\ (\because 2 n-i=n+(n-i)=(n-1)+(n-i \\ +1)=(n-2)+(n-i+2)=\cdots) \text {, so, } \\ T_{\mathrm{n}+1}-T_{\mathrm{n}} \\ =1+1+2+2+3+3+\cdots \text { (up to } n-2 \text { terms), } \\ =\left\{\begin{array}{ll} \frac{1}{4} n(n-2), & \text { ( } n \text { is even }) \\ \frac{1}{4}(n-1)^{2} . & (n \text { is odd) } \end{array}\right. \\ \end{array}$$

Thus, $T_{2 \tau}-T_{2 \mathrm{r}-2}$
$$\begin{array}{l} =\left(T_{2 r}-T_{2 r-1}\right)+\left(T_{2 r-1} \cdots T_{2 r}\right) \\ =(r \cdots 1)^{2}+(r-1)(r-2) \\ =2 r^{2}-5 r+3 . \end{array}$$

Substituting $r$ with $r-1, r-2, \cdots, 4,3$ in (1) and adding all the resulting equations, we get
$$\begin{array}{l} T_{2 \tau}-T_{4}=2 \cdot \frac{1}{6} r(r+1)(2 r+1) \\ -5 \cdot r(r+1)+3 r-\left(2 \times 2^{2}-5 \times 2+1\right) \\ =\frac{1}{6} r(r-1)(4 r-5)-1 . \end{array}$$

Clearly, $T_{4}=1$ (only 2, 3 form a triangle, 1 cannot form a triangle with any two of 2, 3, 4). Therefore,
$$T_{2 r}=\frac{1}{6} r(r-1)(4 r-5) .$$

From this, $T_{2 r+1}=T_{2 r}+r(r-1)$
$$=\frac{1}{6} r(r-1)(4 r+1) \text {, }$$

i.e., $T_{\mathrm{D}}=\left(\begin{array}{l}\frac{1}{24} n(n-2)(2 n-5),(n=2 r) \\ \frac{1}{24}(n-1)(n-3)(2 n-1),\end{array}\right.$
$$(n=2 r+1) \text {. }$$