Note that all integer $x$ work. If $x$ is not an integer then suppose $n<x<n+1$. Then $x=n+\frac{k}{2n+1}$, where $n$ is an integer and $1 \leq k \leq 2n$ is also an integer, since the denominator of the fraction on the right hand side is $2n+1$. We now show that all $x$ of this form work. Note that $x^{2}=n^{2}+\frac{2nk}{2n+1}+\left(\frac{k}{2n+1}\right)^{2}=n^{2}+k-\frac{k}{2n+1}+\left(\frac{k}{2n+1}\right)^{2}$. For $\frac{k}{2n+1}$ between 0 and 1, $-\frac{k}{2n+1}+\left(\frac{k}{2n+1}\right)^{2}$ is between $-\frac{1}{4}$ and 0, so we have $n^{2}+k-1<x^{2} \leq n^{2}+k$, and $\left\lceil x^{2}\right\rceil=n^{2}+k$. Then, $\frac{\left\lceil x^{2}\right\rceil+\lceil x\rceil \cdot\lfloor x\rfloor}{\lceil x\rceil+\lfloor x\rfloor}=\frac{n^{2}+k+n \cdot(n+1)}{2n+1}=n+\frac{k}{2n+1}=x$ so all $x$ of this form work. Now, note that the $2n$ solutions in the interval $(n, n+1)$, together with the solution $n+1$, form an arithmetic progression with $2n+1$ terms and average value $n+\frac{n+1}{2n+1}$. Thus, the sum of the solutions in the interval $(n, n+1]$ is $2n^{2}+2n+1=n^{2}+(n+1)^{2}$. Summing this for $n$ from 0 to 4, we get that the answer is $0^{2}+2\left(1^{2}+2^{2}+3^{2}+4^{2}\right)+5^{2}=85$.