Find all prime numbers $p$ such that the number
$$3^p+4^p+5^p+9^p-98$$
has at most $6$ positive divisors.
To find all prime numbers \( p \) such that the number \( 3^p + 4^p + 5^p + 9^p - 98 \) has at most 6 positive divisors, we proceed as follows:

1. **Check small prime values:**
   - For \( p = 2 \):
     \[
     3^2 + 4^2 + 5^2 + 9^2 - 98 = 9 + 16 + 25 + 81 - 98 = 33
     \]
     The number 33 has the divisors \( 1, 3, 11, 33 \), which are 4 divisors. Hence, \( p = 2 \) works.

   - For \( p = 3 \):
     \[
     3^3 + 4^3 + 5^3 + 9^3 - 98 = 27 + 64 + 125 + 729 - 98 = 847
     \]
     The number 847 has the divisors \( 1, 7, 11, 77, 121, 847 \), which are 6 divisors. Hence, \( p = 3 \) works.

   - For \( p = 5 \):
     \[
     3^5 + 4^5 + 5^5 + 9^5 - 98 = 243 + 1024 + 3125 + 59049 - 98 = 63443
     \]
     The number 63443 has more than 6 divisors. Hence, \( p = 5 \) does not work.

2. **Consider \( p \geq 7 \):**
   - We need to check the congruences modulo 7 and 11:
     \[
     3^p + 4^p \equiv 3 + 4 \equiv 7 \equiv 0 \pmod{7} \quad \text{(since \( p \) is odd)}
     \]
     \[
     5^p + 9^p \equiv 5 + 2 \equiv 7 \equiv 0 \pmod{7} \quad \text{(since \( 9 \equiv 2 \pmod{7} \))}
     \]
     Therefore, \( 3^p + 4^p + 5^p + 9^p \equiv 0 \pmod{7} \).

   - Similarly, modulo 11:
     \[
     3^p + 4^p + 5^p + 9^p \equiv 3 + 4 + 5 + (-2) \equiv 10 \equiv -1 \pmod{11} \quad \text{(since \( 9 \equiv -2 \pmod{11} \))}
     \]
     Therefore, \( 3^p + 4^p + 5^p + 9^p - 98 \equiv -1 - 98 \equiv -99 \equiv 0 \pmod{11} \).

3. **Bounding the number:**
   - Since \( 3^p + 4^p + 5^p + 9^p - 98 \) has at most 6 positive divisors, it must be of the form \( p_1^{e_1} p_2^{e_2} \) where \( p_1 \) and \( p_2 \) are primes and \( e_1, e_2 \leq 2 \). The maximum product of such primes is \( 7 \times 11^2 = 847 \).

4. **Contradiction for \( p \geq 7 \):**
   - For \( p \geq 7 \), the term \( 3^p + 4^p + 5^p + 9^p \) grows very large, much larger than 847. This contradicts the requirement that the number has at most 6 divisors.

Therefore, the only primes \( p \) that satisfy the condition are \( p = 2 \) and \( p = 3 \).

The final answer is \( \boxed{2, 3} \)