Let $\mathbb{A}_n$ be the set containing all such triangles. Without loss of generality, let the three sides of such a triangle be positive integers $a$, $b$, and $c$ satisfying $$a\ge b\ge c > a-b\quad\text{and}\quad\ a+b+c=n$$

When $n=2k$ is even, then it must hold that $$b\le a\le k-1,\quad\text{and}\quad c \ge 2$$

Define mapping $\mathcal{f}$ as $$(a,\ b,\ c)\Rightarrow(a,\ b,\ c-1)$$

Then $\mathcal{f}$ is a mapping from $A_{2k}$ to $A_{2k-1}$, or equivalently, from $A_{2n}$ to $A_{2n-1}$.

It can be shown that $\mathcal{f}$ is an injection, but not a bijection as every element in $\mathbb{A}_{2n}$ has a corresponding one in $\mathbb{A}_{2n-1}$. However, elements in the form of $(a,\ b,\ b)\in\mathbb{A}_{2n-1}$ do not have mappings in $\mathbb{A}_{2n}$. This is because an element $(a,\ b+1,\ b)\in\mathbb{A}_{2n}$ will be mapped to $(a,\ b+1,\ b-1)\in\mathbb{A}_{2n-1}$, not $(a,\ b,\ b)$.

Let $$\mathbb{A}_{2n} - \mathbb{A}_{2n-1}=\mathbb{B}$$

where $\mathbb{B}$ contains all triangles in the form of $(a,\ b,\ b)$ satisfying $$a+2b=2n-1\quad\text{and}\quad b\le a \le 2b-1$$

It follows that $S_{2n}-S_{2n-1}$ equals the number of elements in $\mathbb{B}$, i.e the number of triangles meeting the two conditions above. Solving these two inequalities yields $$\frac{2n-1}{3}\ge b\ge\frac{n}{2}$$

The count of qualifying $b$ will then be $$x= \left\lfloor\frac{2n-1}{3}\right\rfloor - \left\lfloor\frac{n}{2}\right\rfloor + 1$$

Because when $b$ is determined, $a$ will be determined too (as $a+2b=2n-1$), therefore the number of elements in $\mathbb{B}$ is also $x$. Analyzing the modular properties of $n$ will give $$x=\ \left\{ \begin{array}{ll} \left\lfloor\frac{n}{6}\right\rfloor &,\ n\equiv 0,\ 1,\ 3\pmod{6}\\ \\ \left\lfloor\frac{n}{6}\right\rfloor + 1&,\ n\equiv 2,\ 4,\ 5\pmod{6}\end{array}\right.$$

Hence, the claim holds.