I believe $12$ is the largest number with this property, other being $2, 4, 6, 8.$
List. Red are primes $12n-1$, green are primes $12n+1$, factors of $12n-1$ in curly braces, with their sum followed.
The list seems to contain all $a + b = 12n$ primes. Why does 12 have this property and are there more interesting things to know about this?
If we have $$ab \equiv -1\mod 3$$ then there are only two possibilities for $(a,b)$, namely $(1,2)$ or $(2,1)$. Either way we must have $$a+b \equiv 0 \mod 3$$ Similarly, if we have $$ab \equiv -1\mod 4$$ then there are again only two possibilities for $(a,b)$, namely $(1,3)$ or $(3,1)$. So again we must have $$a+b \equiv 0 \mod 4$$ So by combining these we have that if $$ab \equiv -1\mod 12$$ then $$a+b \equiv 0\mod 12$$
So whenever you factor a number $n=12k-1$ into two factors (including $1*n$) their sum will be divisible by $12$.
Note that in general, $$ab \equiv -1\mod m$$ does not lead to the conclusion that $$a+b \equiv 0 \mod m$$ For example, if $m$ is odd and larger than $3$ then $a=2$ and $b=(m-1)/2$ is a counterexample. For $m=16$ a counterexample is $a=3$, $b=5$.
However it does work for $m=8$, in which case the only possibilities are $(1,7)$, $(7,1)$, $(3,5)$, and $(5,3)$. So actually $3*8=24$ is the largest number that has the property that you found.