Inspired by Calvin Lin

We check $a=1,2,3$ and find that only $a=\boxed{3}$ is a solution. (For $a=1$ , the exponent on the LHS, $0^{-1}$ is undefined.)

If $a\geq{4}$ then $a^{(a-1)^{(a-2)}}\geq{a^{(a-1)^2}}>a^{(a-1)(a-2)}=a^{a^2-3a+2}$ .

First, it is tempting to automatically think that $a = 1$ is a solution. However, while the expression on the RHS of the equation becomes $1^{0} = 1,$ the expression on the LHS becomes $\large 1^{0^{-1}},$ which cannot be evaluated since $0^{-1} = \dfrac{1}{0}$ is indeterminate. I'm open to counter-arguments, though, since even if $x$ is indeterminate, $1^{x}$ might still be considered to have a solution of $1.$

Having dealt with, (for the time being), the case of $a = 1,$ for positive integers $a \gt 1$ we will require that the exponents be equal to one another, i.e.,

$(a - 1)^{(a - 2)} = a^{2} - 3a + 2 = (a - 1)(a - 2).$

Now since $(a - 1) \ne 0$ we can divide through by $(a - 1)$ to end up with

$(a - 1)^{(a - 3)} = a - 2.$

Now for a positive integer, raised to an integer power, to yield a value strictly less than the original integer, we must have that power be non-positive. Now since $a \gt 1,$ we can only $(a - 3) \le 0$ when either $a = 2$ or $a = 3.$

For $a = 2$ we have $(2 - 1)^{2 - 3} = 1 \ne 0 = 2 - 2.$

For $a = 3$ we have $(3 - 1)^{3 - 3} = 1 = 3 - 2,$ and thus $a = 3$ is the only possible solution. Checking this value in the original equation, the LHS becomes $\large 3^{2^{1}} = 9,$ and the RHS becomes $\large 3^{(9 - 9 + 2)} = 9,$ thus confirming that $a = 3$ satisfies the original equation.

Since $a = 3$ is the only valid solution, the sum of all positive integer solutions is just $\boxed{3}.$