1. Express the number in base 27:
$$n = 1 \cdot 27^{25} + 2 \cdot 27^{24} + 3 \cdot 27^{23} + \ldots + 25 \cdot 27 + 26$$

2. Rewrite the sum:
$$n = (27^{25} + 27^{24} + \ldots + 27 + 1) + (27^{24} + 27^{23} + \ldots + 27 + 1) + \ldots + (27 + 1) + 1$$

3. Use the geometric series formula:
Each term in the sum can be expressed using the geometric series formula:
$$\sum_{k=0}^{m} 27^k = \frac{27^{m+1} - 1}{26}$$
Applying this to each term, we get:
$$n = \frac{27^{26} - 1}{26} + \frac{27^{25} - 1}{26} + \frac{27^{24} - 1}{26} + \ldots + \frac{27^2 - 1}{26} + \frac{27 - 1}{26}$$

4. Combine the terms:
$$26n = 27^{26} + 27^{25} + 27^{24} + \ldots + 27^2 + 27 - 26$$

5. Use the geometric series formula again:
$$26n = \frac{27^{27} - 1}{26} - 27$$
Simplifying, we get:
$$26n = \frac{27^{27} - 703}{26}$$
Multiplying both sides by 26:
$$676n = 27^{27} - 703$$

6. Simplify using modular arithmetic:
Since $\lambda(100) = 20$ (Carmichael function), we have:
$$3^{81} \equiv 3 \mod 100$$
Therefore:
$$27^{27} \equiv 3^{81} \equiv 3 \mod 100$$
Thus:
$$676n \equiv 3 - 703 \equiv 3 - 703 \equiv -700 \equiv 0 \mod 100$$

7. **Determine possible values of $n \mod 100$:**
Since $676n \equiv 0 \mod 100$ and $\gcd(676, 100) = 4$, we have:
$$676n \equiv 0 \mod 100 \implies n \equiv 0, 25, 50, 75 \mod 100$$

8. Use modulo 4 to narrow down the options:
$$n \equiv (2+4+6+\ldots+26) + 3(1+3+5+\ldots+25) \mod 4$$
The sum of the first series is:
$$2(1+2+3+\ldots+13) = 2 \cdot \frac{13 \cdot 14}{2} = 13 \cdot 14 \equiv 1 \mod 4$$
The sum of the second series is:
$$3(1+3+5+\ldots+25) = 3 \cdot 13^2 = 3 \cdot 169 \equiv 3 \cdot 1 = 3 \mod 4$$
Therefore:
$$n \equiv 1 + 3 = 4 \equiv 0 \mod 4$$
This eliminates 25, 50, and 75, leaving:
$$n \equiv 0 \mod 100$$

The final answer is $\boxed{0}$