Can a Red Black tree be constructed of only black nodes using RB insert only?
Properties of Red Black Tree:

0) Every node is black or red. Ok, no problem.

1) Root is black. Ok, no problem.

2) All leaves (empty nodes) are black. Ok, no problem.

3) Every red node must have two black childs. Ok, no problem.

4) Every path from any node to leaf has equal number of black nodes. It seems problematic.

So now comes problems: When you insert element it's color is red. (In fact this not comes from properties but observation that adding red node does not violate $4$th rule, so it is easier to implement).

So for start you add one element, let me say 10.
It is black root. Ok.

Now You add 13. It must be red, otherwise it violates rule 4.



Now you add 19. Depending on implementation you can leave all black, or recolor.



Now adding 12 gives recoloring and comes red.



And after several numbers



And now it starts nightmare, as you should propagate changes from one side of tree to another convincing somehow that you will change everything to black after insertion.

Basically I asked you about rules to obey, as it comes to mind that with $4$th can be fulfilled only in perfectly balanced trees.

So to answer your questions:

No you cannot with common implementations of RBT, but extending balance operation to recolor when it is possible - it does not violate rules, so it is possible. Also with extension that whenever there is $n = 2^k - 1$ nodes, restructure it to perfectly balanced tree and recolor everything to black, it would be also possible.

I did firstly answer "no" as purpose of balancing operation is to guarantee bound on worst case search / insert, and such extensions comes at price of time complexity. It does not violate rules, but is not implicitly implemented in standard RBT.

Given only root, you have it. Given three nodes inserted in order with implementation changing all to black (rather rare case and counterproductive), you can. But this is last one that is without deletions possible.

When you are about to insert element and want all tree to be black that would mean there is violated rule $4$ (so not possible) or you have to rebalance tree, which would recolor nodes, so it is not pure black anymore.

With deletions on the other hand ;)