4. Let $a, b$ be positive integers not exceeding 12, satisfying: there exists a constant $C$, such that $a^{n}+b^{n+9} \equiv C(\bmod 13)$ for any positive integer $n$. Find all ordered pairs $(a, b)$ that satisfy the condition.
Given the condition, for any positive integer $n$, we have $a^{n}+b^{n+9} \equiv a^{n+3}+b^{n+12}(\bmod 13)$. Note that 13 is a prime number, and $a, b$ are both coprime with 13. By Fermat's Little Theorem, we know $a^{12} \equiv b^{12} \equiv 1(\bmod 13)$. Taking $n=12$ in (1), we simplify to get $1+b^{9} \equiv a^{3}+1(\bmod 13) \Rightarrow b^{9} \equiv a^{3}(\bmod 13)$. Substituting into (1) gives $a^{n}+a^{3} b^{n} \equiv a^{n+3}+b^{n}(\bmod 13) \Rightarrow\left(a^{n}-b^{n}\right)\left(a^{3}-1\right) \equiv 0(\bmod 13)$. 

Case 1: If $a^{3} \equiv 1(\bmod 13) \Rightarrow b^{3} \equiv a^{3} b^{3} \equiv b^{9} b^{3} \equiv 1(\bmod 13)$, and $a, b \in\{1,2, \cdots, 12\}$. After inspection, we find $a, b \in\{1,3,9\}$.
In this case, $a^{n}+b^{n+9} \equiv a^{n}+b^{n} \equiv C(\bmod 13)$ holds for any positive integer $n$, thus $a+b \equiv a^{3}+b^{3} \equiv 2 \Rightarrow a=b=1$.

Upon inspection, when $(a, b)=(1,1)$, for any positive integer $n$, $a^{n}+b^{n+9}$ is congruent to 2 modulo 13, which is a constant, satisfying the condition;
Case 2: If $a^{3} \not \equiv 1(\bmod 13)$, then for any positive integer $n$, we have $a^{n} \equiv b^{n}(\bmod 13)$, particularly, $a \equiv b(\bmod 13) \Rightarrow a=b$.
Thus, $a^{3} \equiv b^{9} \equiv a^{9}(\bmod 13) \Rightarrow a^{3}\left(a^{3}-1\right)\left(a^{3}+1\right) \equiv 0(\bmod 13)$ $\Rightarrow a^{3} \equiv-1(\bmod 13)$, hence $a \equiv-1,-3,-9(\bmod 13) \Rightarrow a=4,10,12$. Upon inspection, when $(a, b)=(4,4),(10,10),(12,12)$, for any positive integer $n$, $a^{n}+b^{n+9} \equiv a^{n}+a^{n+9} \equiv a^{n}\left(1+a^{9}\right) \equiv 0(\bmod 13)$, satisfying the condition. In summary, all ordered integer pairs that satisfy the condition are $(1,1),(4,4),(10,10),(12,12)$.