Answer: b) there are an infinite number of nice numbers.

a) Let $N$ be a nice number, i.e. $N = d_1^4 + d_2^4 + d_3^4 + d_4^4 + d_5^4$, where $d_i$, $i = 1, 2, 3, 4, 5$, are the distinct divisors of $N$. If some divisor of $N$ is divisible by $5$, then $N$ is divisible by $5$.
So we suppose that $d_1, d_2, d_3, d_4, d_5$ are not divisible by $5$. Then their fourth powers are congruent to $1$ modulo $5$. Indeed, if $M = 5k \pm 1$ ($k \in \mathbb{N}$), then $M^2 = 25k^2 \pm 10k + 1 = 5m + 1$ ($m \in \mathbb{N}$) and $M^4 = (5m+1)^2 = 25m^2 + 10m + 1$, i.e. $M^4$ is congruent to $1$ modulo $5$; if $M = 5k \pm 2$ ($k \in \mathbb{N}$), then $M^2 = 25k^2 \pm 20k + 4 = 5m + 4$ ($m \in \mathbb{N}$) and $M^4 = (5m+4)^2 = 25m^2 + 40m + 16$, i.e. $M^4$ is congruent to $1$ modulo $5$. Therefore, the fourth power of any divisor $d_i$ is congruent to $1$ modulo $5$, i.e. $d_i^4 = 5k_i + 1$, and it follows that $N$ is divisible by $5$.

b) Let there exist a nice number $N$, i.e. $N = d_1^4 + d_2^4 + d_3^4 + d_4^4 + d_5^4$, where $d_i$ are some positive integers and $p > 1$. If $d_1, d_2, d_3, d_4, d_5$ are distinct divisors of $N$, consider the number $N(p) = Np^4$, where $p$ is any positive integer greater than $1$. It is obvious that $d_1p, d_2p, d_3p, d_4p, d_5p$ are distinct divisors of $N(p)$, and
$$\begin{aligned} N(p) &= Np^4 = [N = d_1^4 + d_2^4 + d_3^4 + d_4^4 + d_5^4] = (d_1^4 + d_2^4 + d_3^4 + d_4^4 + d_5^4)p^4 = \\ &= (d_1p)^4 + (d_2p)^4 + (d_3p)^4 + (d_4p)^4 + (d_5p)^4. \end{aligned}$$
So there are infinitely many nice numbers $N(p)$ if there exists at least one nice number. It remains to note that the number $N = 1^4 + 2^4 + 3^4 + 6^4 + 34^4$ is nice. Indeed, it is evident that $N$ is divisible by $2$. Since the numbers $1^4, 2^4$, and $34^4$ are congruent $1$ modulo $3$, the number $N$ is divisible by $3$. Since $N$ is even and divisible by $3$, we see that $N$ is divisible by $6$. Besides,
$$N = 1^4 + 2^4 + 3^4(1^4 + 2^4) + 34^4 = (1^4 + 2^4)(1^4 + 3^4) + 34^4 = 17 \cdot (1^4 + 3^4) + 34^4,$$
whence we see that $N$ is divisible by $34$. Thus, $N$ is nice.