Claim: There exists such a chain, beginning in a corner, in a triangle of size n for any natural number n.
Proof: Induction on n.
Base case: For n = 1, the triangle itself is a chain of length 1 = 1^2 - 1 + 1.
Induction step: Pick a side adjacent to the given corner. The triangles with at least one vertex on that side form a chain of length 2n - 1. Take all the elements of that chain except the final one. The triangles not in that chain form a triangle of side n-1, of which you are now adjacent to the corner. By the induction hypothesis, then, we may extend the chain by a further (n-1)^2 - (n-1) + 1 triangles giving a total length of n^2 - 2n + 1 - n + 1 + 1 + 2n - 2 = n^2 - n + 1.

Of course, this proof does not give an explicit construction, but it'suggests two reasonably pleasant ones: a zig-zag and a spiral.

Thankyou

sfwc
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Thinking about a proof, 2(k-1) triangles on a level of side k form a parallelogram such that, when chained, the exit and the entry points of the chain are in the opposite corners. This allows stacking the levels on top of each other. The total number of triangles in the chain levels is then

2(n-1) + 2(n-2) + ... + 2 = n(n-1),

to which one should add the top triangle - a level of its own.