The $n$ that work are $n=2, 3, 4$. $n=1$ clearly doesn't work. Claim: $n\geq5$ does not work. Pick $a=1, b=\varphi, c=\varphi^2$. Let's look at max$(k, l, m)$. If all of them are the same it clearly does not work. If $m$ is the maximum, then $cn^m = bn^m+an^m >bn^l+an^k$ contradiction. If $k$ is the lone maximum, then $an^k \geq 5an^{k-1} > bn^{k-1}+cn^{k-1} \geq bn^l+cn^m$ contradiction. If $k=l$ and they are both the maximum, $bn^k-an^k \geq 5(bn^{k-1}-an^{k-1}) > cn^{k-1} \geq cn^m$ contradiction. For $n \leq 4$, we will prove it is possible. First, use the powers of $n$ to make the ratio between any 2 of $an^k, bn^l, cn^m$ be between 1/4 and 4. Let's pick those numbers to be our new $a, b, c$ and assume WLOG $a\leq b \leq c$ and $a=1$. Assume FTSOC that no $(k, l, m)$ will make them be sides of a triangle. So $4 \geq b+c$, $c \geq b+1$, $4b \geq c+4$. $4b \geq c+4 \geq b+5$ so $b \geq 5/3$ so $c \geq 8/3$ so $4 \geq b+c \geq 13/3 > 4$ contradiction.