All fantastic sequences are: $x_k = a(k \log(k) - (k-1) \log(k-1)) + c$, where $a$ is a nonnegative real number, and $c$ is an arbitrary real number.

First it is easy to see that if $\{x_1, x_2, \cdots, x_k, \cdots\}$ is fantastic, then for any real number $c$ and positive real number $r$, $\{\frac{x_1}{r} + c, \frac{x_2}{r} + c, \cdots, \frac{x_k}{r} + c, \cdots\}$ is also fantastic.
Therefore, without loss of generality assume $x_1 = 0$. Let $f(p) = \sum_{i=1}^{p} x_i$, then it is easy to see that $H(\frac{q}{p}) = \frac{f(p)-f(q)-f(p-q)}{p}$.
Since $H(\frac{q}{p}) = H(\frac{kp}{kp})$, therefore
$$\frac{f(p) - f(q) - f(p-q)}{p} = \frac{f(kp) - f(kq) - f(k(p-q))}{kp}$$
After rearranging terms and simplifying we obtain
$$(f(kp) - k f(p)) = (f(kq) - k f(q)) + (f(k(p-q)) - k f(p-q))$$
which holds for all positive integers $k, p$ and non-negative integers $0 \le q \le p$. Now fix a positive integer $k$, and define $g_k(p) = f(kp) - kf(p)$, then the above formula becomes
$$g_k(p) = g_k(q) + g_k(p - q)$$
This is a standard Cauchy equation; setting $q=1$ and using mathematical induction, we get $g_k(p) = pg_k(1) = p(f(k) - kf(1)) = pf(k)$ (since $f(1) = x_1 = 0$), therefore
$$f(kp) = kf(p) + pf(k)$$
Furthermore, if we fix $p$ in this formula and use mathematical induction, we can obtain that for any positive integers $p, k$, $f(p^k) = p^{k-1}kf(p)$.

For two positive integers $q < p$, since $\frac{q}{p+q+1} < \frac{q+1}{p+q+1} \le \frac{1}{2}$ and by the increasing property of $H(x)$, we have
$$\begin{aligned} \frac{f(p+q+1) - f(q) - f(p+1)}{p+q+1} \le \frac{f(p+q+1) - f(q+1) - f(p)}{p+q+1} \end{aligned}$$
After rearranging we obtain
$$f(p+1) - f(p) \ge f(q+1) - f(q)$$
which holds for all $p > q$. Note that this means $f(p)$ is a convex function.
From this, we can obtain an obvious inequality: for positive integers $p > q > r$,
$$\frac{f(p) - f(q)}{p - q} \ge \frac{f(p) - f(r)}{p - r} \ge \frac{f(q) - f(r)}{q - r}$$
This is because
$$\begin{aligned} & \frac{f(p) - f(q)}{p - q} \\ \ge & \frac{(p - q)(f(q + 1) - f(q))}{p - q} \\ = & f(q + 1) - f(q) \\ = & \frac{(q - r)(f(q + 1) - f(q))}{q - r} \\ = & \frac{f(q) - f(r)}{q - r} \end{aligned}$$
And since
$$\frac{f(p) - f(r)}{p - r} = \frac{p - q}{p - r} \frac{f(p) - f(q)}{p - q} + \frac{q - r}{p - r} \frac{f(q) - f(r)}{q - r}$$
therefore its value lies between the two.

Now suppose $f(2) = a$ is known, then $a = f(2) = \frac{f(2)-f(1)-f(1)}{2} = H(\frac{1}{2}) \ge H(0) = 0$. According to $f(p^k) = p^{k-1}kf(p)$, we can deduce $f(2^k) = 2^{k-1}ka$, that is, when $p = 2^k$, $f(p) = \frac{a}{2}p \log 2p$. When $p \neq 2^k$, we hope to estimate the value of $f(p)$ using inequalities. For any positive integer $m$, there exists a positive integer $n$ such that $2^{n+1} > p^m > 2^n$, therefore
$$\frac{f(2^{n+1}) - f(2^n)}{2^{n+1} - 2^n} \ge \frac{f(p^m) - f(2^n)}{p^m - 2^n}$$
After rearranging and substituting $f(2^k) = 2^{k-1}ka$ we obtain
$$a(2^{n-1}n + (\frac{n}{2} + 1)(p^m - 2^n)) \geq f(p^m) = p^{m-1}mf(p)$$
Further we have
$$f(p) \le \frac{apn + 2 - \frac{2^n}{p^m}}{2m} \le \frac{apn + 2}{2m}$$
From $2^{n+1} > p^m > 2^n$ we can deduce $\frac{n+1}{m} > \log_2 p > \frac{n}{m}$, so $\frac{n}{m}$ is a rational estimate of $\log_2 p$. Therefore
$$f(p) \le \frac{apn + 2}{2m} \le \frac{ap}{2}(\log_2 p + \frac{2}{m})$$
When $m$ is very large we obtain
$$f(p) \leq \frac{ap}{2} \log_2 p$$
Similarly, using $p^m > 2^n > 2^{n-1}$ we can obtain
$$\frac{f(p^m) - f(2^n)}{p^m - 2^n} \geq \frac{f(2^n) - f(2^{n-1})}{2^n - 2^{n-1}}$$
After rearranging we obtain
$$f(p) \geq \frac{ap}{2} \frac{n + 1 - \frac{2^n}{p^m}}{m} \geq \frac{ap}{2} (\log_2 p - \frac{1}{m})$$
When $m$ is very large we then have
$$f(p) \geq \frac{ap}{2} \log_2 p$$
Combining the two inequality estimates together we obtain
$$f(p) = \frac{ap}{2} \log_2 p$$
Combining $\frac{a}{2\log_2 p}$ into a single parameter, and combining with the case $p = 2^k$, we obtain
$$f(p) = ap \log p$$
which holds for any positive integer, where $a$ is a nonnegative real number. Therefore
$$x_k = f(k) - f(k-1) = a(k \log(k) - (k-1) \log(k-1))$$