Let $T_n=\{11(k+h)+10(n^k+n^h)\mid1\leq k,h \leq 10\}$ for $n\in\mathbb N$. Find all $n$ such that there is no $a\neq b\in T_n$ with $110 \mid (a-b)$. With every positive integer $n$, we have a set $T_n=\{11(k+h)+10(n^k+n^h) \mid 1\leq k,h \leq 10\}$. Find all $n$ such that there is no $a\neq b\in T_n$ with $110 \mid (a-b)$.
I tried to divided it into some case with $n\leq 11$, and then to discard the case not right. Is there some other way to solve it?

We can solve this without resorting to brute force.

Firstly, define, for positive integer $n$ and $1\leq k,h\leq10$, $f_n(k,h)=11(k+h)+10(n^k+n^h)$. Then the condition $\exists a,b\in T_n$ such that $a\ne b$ and $a\equiv b\pmod{110}$ is equivalent with the conditions below.
$$\begin{cases}
n^{k_1}+n^{h_1}&\equiv n^{k_2}+n^{h_2}\pmod{11}\\
k_1+h_1&\equiv k_2+h_2\pmod{10}\\
f_n(k_1,h_1)&\ne f_n(k_2,h_2)
\end{cases}.$$
If $(k_1,h_1,k_2,h_2)$ satisfies the above congruences, then call it a valid pair.

If $n^5\equiv1\pmod{11}$, then $n^1\equiv n^6\pmod{11}$ and $n^2\equiv n^7\pmod{11}$. Also $1+2\equiv6+7\pmod{10}$, so such an $n$ does not satisfy the conditions. By the same token, if $n\equiv-1\pmod{11}$ then $n$ does not satisfy the condition. Hence all that rest are those $n$ such that if $n^k\equiv1\pmod{11}$ then $10\mid k$, i.e. primitive roots of $11$, which are $\equiv2,6,7,8\pmod{11}$.
So let $n$ be a primitive root modulo $11$.
Then the condition $n^{k_1}+n^{h_1}\equiv n^{k_2}+n^{h_2}\pmod{11}$ becomes 
$$n^{k_1}+n^{h_1}\equiv n^{k_2}+n^{k_1+h_1-k_2}\pmod{11}.$$
Hence $n^{k_2}$ satisfies the equation $$x^2-(n^{k_1}+n^{h_1})x+n^{k_1+h_1}\equiv0\pmod{11}.$$
But $x^2-(n^{k_1}+n^{h_1})x+n^{k_1+h_1}=(x-n^{k_1})(x-n^{h_1})$, thus $n^{k_2}$ is congruent modulo $11$ to either $n^{k_1}$ or $n^{h_1}$. Since $n$ is a primitive root modulo $11$, this implies that $k_2=k_1$ or $k_2=h_1$. This shows that $f_n(k_1,h_1)=f_n(k_2,h_2)$, and hence $(k_1,h_1,k_2,h_2)$ is not a valid pair.
Therefore indeed for $n\equiv2,6,7,8\pmod{11}$, there are no valid pairs, i.e. such $n$ are what we are looking for.

Hope this helps.
