Step 1: Write a formula for $T_n$.
To form a triangle, we choose 3 vertices from the $n$ vertices of the regular polygon. No three vertices of a regular polygon are collinear (for $n \geq 3$), so any 3 chosen vertices always form a valid triangle. Thus $T_n = \binom{n}{3}$.
Step 2: Write $T_{n+1}$ similarly.
For an $(n+1)$-sided polygon: $T_{n+1} = \binom{n+1}{3}$.
Step 3: Simplify $T_{n+1} - T_n$ using Pascal triangle identity.
The Pascal identity states $\binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1}$ for any $r$. Applying with $r=3$: $\binom{n+1}{3} = \binom{n}{3} + \binom{n}{2}$. Therefore $T_{n+1} - T_n = \binom{n}{2}$.
Step 4: Set up the equation from the given condition.
Given $T_{n+1}-T_n=10$, we have $\binom{n}{2}=10$, i.e., $\frac{n(n-1)}{2}=10$, so $n(n-1)=20$.
Step 5: Solve the quadratic.
$n^2-n-20=0$. Factoring: $(n-5)(n+4)=0$, giving $n=5$ or $n=-4$. Since $n$ must be a positive integer $\geq 3$, we take $n=5$.
Step 6: Verify and state the final answer.
Check: $T_5 = \binom{5}{3}=10$, $T_6=\binom{6}{3}=20$, $T_6-T_5=10$. \[ \boxed{n=5} \]