a) Let $N$ be a nice number, i.e. $N = d_1^2 + d_2^2 + d_3^2$, where $d_1, d_2, d_3$ are distinct divisors of $N$. If some divisor of $N$ is divisible by $3$, then $N$ is divisible by $3$.

So we suppose that $d_1, d_2, d_3$ are not divisible by $3$. Then their squares are congruent to $1$ modulo $3$, i.e. $d_1^2 = 3k_1 + 1$, $k_1 \in \mathbb{N}$, $i = 1, 2, 3$. Therefore
$$N = d_1^2 + d_2^2 + d_3^2 = 3(k_1 + k_2 + k_3) + 3,$$
and so $N$ is divisible by $3$.

b) There exists a nice number $N'$. For example, if $N = 30$ and its divisors are $d_1 = 1$, $d_2 = 2$, $d_3 = 5$, then
$$d_1^2 + d_2^2 + d_3^2 = 1^2 + 2^2 + 5^2 = 1 + 4 + 25 = 30 = N,$$
i.e. $N$ is nice.

Consider $N(p) = Np^2$, where $p$ is some positive integer and $p > 1$. If $d_1, d_2, d_3$ are distinct divisors of $N$, then it is obvious that $d_1p, d_2p, d_3p$ are distinct divisors of $N(p)$, and
$$N(p) = Np^2 = [N = d_1^2 + d_2^2 + d_3^2] = (d_1^2 + d_2^2 + d_3^2)p^2 = (d_1p)^2 + (d_2p)^2 + (d_3p)^2.$$
From this equality it follows that $N(p)$ is nice too. Since any positive integer can be used as $p$, there are infinitely many nice numbers $N(p)$.