1. Given Conditions:
- We need to prove that for all positive integers $n$, $169$ divides $21n^2 + 89n + 44$ if $13$ divides $n^2 + 3n + 51$.

2. Condition Analysis:
- If $13$ divides $n^2 + 3n + 51$, then $n^2 + 3n + 51 \equiv 0 \pmod{13}$.

3. Simplify the Congruence:
- We need to find $n$ such that $n^2 + 3n + 51 \equiv 0 \pmod{13}$.
- Simplify $51 \mod 13$:
$$51 \div 13 = 3 \quad \text{remainder} \quad 12 \quad \Rightarrow \quad 51 \equiv 12 \pmod{13}$$
- Thus, the congruence becomes:
$$n^2 + 3n + 12 \equiv 0 \pmod{13}$$

4. Solve the Quadratic Congruence:
- We solve $n^2 + 3n + 12 \equiv 0 \pmod{13}$.
- The roots of the quadratic equation $n^2 + 3n + 12 = 0$ can be found using the quadratic formula:
$$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a = 1$, $b = 3$, and $c = 12$.
- Calculate the discriminant:
$$\Delta = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 12 = 9 - 48 = -39$$
- Since we are working modulo $13$:
$$-39 \equiv 10 \pmod{13}$$
- The discriminant modulo $13$ is $10$, and we need to find the square root of $10$ modulo $13$. By inspection or using quadratic reciprocity, we find that $10$ is a quadratic residue modulo $13$, and its square roots are $6$ and $7$ (since $6^2 \equiv 36 \equiv 10 \pmod{13}$ and $7^2 \equiv 49 \equiv 10 \pmod{13}$).

5. Find the Roots:
- Using the roots $6$ and $7$:
$$n \equiv \frac{-3 \pm 6}{2} \pmod{13} \quad \text{and} \quad n \equiv \frac{-3 \pm 7}{2} \pmod{13}$$
- Simplify the roots:
$$n \equiv \frac{3}{2} \pmod{13} \quad \text{and} \quad n \equiv \frac{-9}{2} \pmod{13}$$
- Since $2$ has an inverse modulo $13$ (which is $7$ because $2 \cdot 7 \equiv 1 \pmod{13}$):
$$n \equiv 3 \cdot 7 \equiv 21 \equiv 8 \pmod{13} \quad \text{and} \quad n \equiv -9 \cdot 7 \equiv -63 \equiv 10 \pmod{13}$$
- Thus, $n \equiv 8 \pmod{13}$ or $n \equiv 10 \pmod{13}$.

6. **General Form of $n$:**
- Let $n = 13k + 8$ or $n = 13k + 10$ for some integer $k$.

7. **Substitute $n$ into $21n^2 + 89n + 44$:**
- For $n = 13k + 8$:
$$21(13k + 8)^2 + 89(13k + 8) + 44$$
- Expand and simplify:
$$21(169k^2 + 208k + 64) + 89(13k + 8) + 44$$
$$= 21 \cdot 169k^2 + 21 \cdot 208k + 21 \cdot 64 + 89 \cdot 13k + 89 \cdot 8 + 44$$
$$= 21 \cdot 169k^2 + 21 \cdot 208k + 1344 + 1157k + 712 + 44$$
$$= 21 \cdot 169k^2 + (21 \cdot 208 + 1157)k + (1344 + 712 + 44)$$
$$= 21 \cdot 169k^2 + 169 \cdot 23k + 2100$$
$$= 169(21k^2 + 23k + 12)$$
- Clearly, $169$ divides $21n^2 + 89n + 44$.

8. Conclusion:
- The same steps can be applied for $n = 13k + 10$ and will yield a similar result.

$\blacksquare$