It's relatively simple to prove that this sum converges by using that $|\sin(t)|\leq|t|$ to get that $|f(x)|<\frac{\pi^2 x}{6}$, but this doesn't give boundedness.
I've tried using Abel summation and the approximation $\lfloor t \rfloor \approx t$, but the maximum possible error term is linear in $x$ ($2x/\pi$ if I remember correctly), and I've tried to use the Chinese Remainder Theorem to construct values of $x=\frac{\pi}{2} N$ for which the $\sin(x/n)\approx 1$ for small values of $n$, but I haven't been able to get anywhere with that as the behavior of numbers that are close to $n\bmod 4n$ for lots of $n$ is not very nice. My hunch says that there should be values of $x$ for which "the stars align" and the first few terms are close to $1/n$, where "few" is enough for $f(x)$ to get arbitrarily big, but I'm not sure if there's a way to rigorize this via a probabilistic argument.
For completeness, here's a bit of what I've done so far:
Say it were possible for $\sin(x/n)$ to equal $1$ for every $1\leq n\leq K$. Then
$$f(x) = H_K+\sum_{n=K+1}^{\lfloor x\rfloor} \frac{\sin(x/n)}{n} + \sum_{n=\lfloor x\rfloor+1}^{\infty} \frac{\sin(x/n)}{n}.$$
Using for the first sum that $-1\leq \sin(x)$ and for the second sum that $t\sin(1)\leq \sin(t)$ for $0\leq t\leq 1$,
$$f(x)\geq H_K - \left(H_{\lfloor x\rfloor}-H_{K+1}\right)+x\sin(1)\zeta\left(2,\lfloor x\rfloor+1\right),$$
where $\zeta(s,q)$ is the Hurwitz Zeta Function. It shouldn't be too hard to prove that $\zeta(2,q)>1/q$ (although I don't know of a proof offhand), so this implies
$$f(x)\geq H_K - \left(H_{\lfloor x\rfloor}-H_{K+1}\right)+\frac{\sin(1)x}{\lfloor x\rfloor+1}.$$
If $K$ and $x$ are both *large*, this is asymptotically about
$$\ln\left(\frac{K(K+1)}{x}\right)+C$$
for some constant $C$. Thus, if we could make $K$ grow a bit faster than $\sqrt{x}$, we would be able to prove that $f$ is unbounded.
Unfortunately, $\sin(x/n)=1$ for all $1\leq n\leq K$ is not possible for $K\geq 2$ for modular reasons (specifically numbers that are $1\bmod 4$ are not also $2\bmod 8$). So, we would need to find a threshold $\mu$ as well as a $K$ for which $\sin(x/n)>\mu$ for all $1\leq n\leq K$, and then weaken our bounds accordingly. Or, we could try to use a better bound than $-1$ on $\sin$ for the first sum. I'm not really sure.
