To solve the problem, we need to determine all positive integers that are relatively prime to every term of the sequence defined by:

$$a_n = 2^n + 3^n + 6^n - 1, \quad n \geq 1.$$

Step 1: Understanding the sequence properties

To determine an integer relatively prime to all $a_n$, we first investigate the properties of the sequence:

$$a_n = 2^n + 3^n + 6^n - 1 = 2^n + 3^n + (2 \cdot 3)^n - 1.$$

Step 2: Checking divisibility by small primes

Let's check the sequence for small integer divisibility patterns, beginning with the smallest prime number, $p = 2$:

- For $n = 1$:
$$a_1 = 2^1 + 3^1 + 6^1 - 1 = 2 + 3 + 6 - 1 = 10$$
$a_1$ is divisible by $2$.

- For $n = 2$:
$$a_2 = 2^2 + 3^2 + 6^2 - 1 = 4 + 9 + 36 - 1 = 48$$
$a_2$ is divisible by $2$.

- In general, if we use modulo 2 for any $n \geq 1$, it is evident that $a_n \equiv 0 \pmod{2}$.

Similarly, let's check for divisibility by $3$:

- For $n = 1$:
$$a_1 = 10 \equiv 1 \pmod{3}$$
$a_1$ is not divisible by $3$.

- For $n = 2$:
$$a_2 = 48 \equiv 0 \pmod{3}$$
$a_2$ is divisible by $3$.

- For $n = 3$:
$$a_3 = 2^3 + 3^3 + 6^3 - 1 = 8 + 27 + 216 - 1 = 250 \equiv 1 \pmod{3}$$
$a_3$ is not divisible by $3$.

Notice that because $a_2 \equiv 0 \pmod{3}$, this implies $a_n$ shares periodic divisibility by $3$.

Conclusion

Through examining divisibility by smaller primes such as $2$ and $3$, and recognizing these properties, we deduce that the only positive integer that is relatively prime to every $a_n$ is:

$$\boxed{1}$$
This is because $1$ is relatively prime to every integer. Hence, the complete set of integers relatively prime to all terms in the sequence is \{1\}, given their universal property.