Suppose that $n-3=2a+3b+4c$, where $a,b$, and $c$ are non-negative integers. We may think of this as a partition of $n-3$ into parts taken from $\{2,3,4\}$. If $p_1=p_2=a+b+c$, $p_3=b+c$, and $p_4=c$, the conjugate partition is $n-3=p_1+p_2+p_3+p_4$, with the proviso that $p_3$ and $p_4$ can be $0$.
Let
$$\begin{align*}
r&=p_1+p_4+1=a+b+2c+1\;,\\
s&=p_2+1=a+b+c+1\;,\text{ and}\\
t&=p_3+1=b+c+1\;;
\end{align*}$$
clearly $r+s+t=n$, and $r\ge s\ge t\ge 1$. Moreover, $s+t-r=b+1>0$, so there is a non-degenerate triangle with sides $r,s$, and $t$ and perimeter $n$.
Now suppose that $r,s$, and $t$ are the sides of a non-degenerate triangle with perimeter $n$, where $r\ge s\ge t$. Let
$$\begin{align*}
p_1&=s-1\;,\\
p_2&=s-1\;,\\
p_3&=t-1\;,\text{ and}\\
p_4&=r-s\;.
\end{align*}$$
Clearly $p_1+p_2+p_3+p_4=r+s+t-3=n-3$, and $p_1=p_2\ge p_3$. Moreover, if $p_3
$$\begin{align*}
c&=p_4\;,\\
b&=p_3-p_4\;,\text{ and}\\
a&=p_2-p_3\;;
\end{align*}$$
then
$$2a+3b+4c=2p_2+p_3+p_4=p_1+p_2+p_3+p_4=n-3\;.$$
This establishes the desired bijection.
By the way, this is OEIS A005044, with several interesting references; I especially liked this PDF and this one.
