$T_n=\frac{n(n-1)}{2}$ and $4T_n=2n(n-1)=2n^2-2n$
And $\gcd(n-2, 2n^2-2n)=\gcd(n-2, 2n^2-2n-2n(n-2))$ by Euclidean algorithm.
And $2n^2-2n-2n(n-2)=2n^2-2n-[2n^2-4n]=2n$, so it remains to maximize $\gcd(n-2, 2n)=\gcd(n-2, 2n-2(n-2))=\gcd(n-2, 4)$ which has maximum value of $\boxed{4}$.
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