If $y=x+k$ then $\frac{x\cdot d}{y}<d$, so $\left\lceil\frac{x\cdot d}{y}\right\rceil\cdot \frac{x}{d}\leq x<y<\left\lceil\left(\sqrt{y}+1\right)^2\right\rceil$, contradiction. Therefore $x=y+k$. $y+2+2\sqrt{y}>\left\lceil\left(\sqrt{y}+1\right)^2\right\rceil=\left\lceil\frac{x\cdot d}{y}\right\rceil\cdot \frac{x}{d}\geq \frac{x^2}{y}=y+2k+\frac{k^2}{y}>y+2k$, so $y>(k-1)^2$ Now $\frac{x\cdot d}{y}=d+\frac{k\cdot d}{y}$, so $\left\lceil\frac{x\cdot d}{y}\right\rceil\cdot \frac{x}{d}=x+\left\lceil\frac{k\cdot d}{y}\right\rceil\cdot \frac{x}{d}$ Meanwhile, $\left\lceil\left(\sqrt{y}+1\right)^2\right\rceil=y+1+\lceil 2\sqrt{y}\rceil$, so the equation becomes \[k+\left\lceil\frac{k\cdot d}{y}\right\rceil\cdot \frac{x}{d}=1+\lceil 2\sqrt{y}\rceil\]Case 1. $d=k$ Then $d$ divides $y$. Let $y=ky'$. Since $y>(k-1)^2$, $y'\geq k-1$. If $y'\geq k$, then $\left\lceil\frac{k\cdot d}{y}\right\rceil=1$, so $k+y'+1=1+\lceil 2\sqrt{ky'}\rceil$ We can easily check that $y'$ such that $k\leq y'<k+2\sqrt{k}+1$ is possible, so there are $\lceil 2\sqrt{k}+1\rceil$ solutions. If $y'=k-1$, then $\left\lceil\frac{k\cdot d}{y}\right\rceil=2$, so the equation becomes $3k=2k$, which is a contradiction. Case 2. $d<k$ Case 2-1. k=2,d=1 Then the equation becomes $2+\left\lceil\frac{2}{y}\right\rceil\cdot(y+2)=1+\lceil 2\sqrt{y}\rceil$. We can easily see that there are no solutions. Case 2-2. Otherwise... Since $d$ is a divisor of $k$, $d\leq \frac{k}{2}$, so $d\leq k-2$. Therefore $y>(k-1)^2>k(k-2)\geq kd$, so the equation becomes $k+\frac{k+y}{d}=1+\lceil 2\sqrt{y}\rceil$. Since $y>k(k-2)$, $\frac{y}{d}\geq \frac{y}{k-2}\geq \frac{y}{k}+2$. Also, $\frac{k}{d}\geq 1$. $\therefore k+\frac{k+y}{d}\geq 3+k+\frac{y}{k}\geq 3+2\sqrt{y}>1+\lceil 2\sqrt{y}\rceil$. A contradiction! From all cases, we get $S(k)=\lceil 2\sqrt{k}+1\rceil$. $\blacksquare$