Round-up and round-down

If $n$ is a perfect square, say $n=m^2$ , then $f(m^2)=g(m^2)=m$ . If this is not the case, set $n=m^2+k$ , where $0<k<2m+1$ . We then have:

$f(m^2+k)=\left\lfloor\dfrac{m^2+k}{\left\lfloor\sqrt{m^2+k}\right\rfloor}\right\rfloor=\left\lfloor\dfrac{m^2+k}{m}\right\rfloor=m+\left\lfloor\dfrac{k}{m}\right\rfloor\\ g(m^2+k)=\left\lceil\dfrac{m^2+k}{\left\lceil\sqrt{m^2+k}\right\rceil}\right\rceil=\left\lceil\dfrac{m^2+k}{m+1}\right\rceil=m-1+\left\lceil\dfrac{k+1}{m+1}\right\rceil$

• If $k<m$ , then $\left\lfloor\dfrac{k}{m}\right\rfloor=0$ and $\left\lceil\dfrac{k+1}{m+1}\right\rceil=1$ , then we get that $f(m^2+k)=g(m^2+k)=m$ .
• If $k=m$ , then $\left\lfloor\dfrac{k}{m}\right\rfloor=\left\lceil\dfrac{k+1}{m+1}\right\rceil=1$ , which implies $f(m^2+k)=m+1$ and $g(m^2+k)=m$ .
• If $m<k<2m$ , then we have $1<\dfrac{k}{m}<2$ , so that $\left\lfloor\dfrac{k}{m}\right\rfloor=1$ and $f(m^2+k)=m+1$ ; this also implies $m+1<k+1<2(m+1)$ , implying $1<\dfrac{k+1}{m+1}<2$ . Thus, $\left\lceil\dfrac{k+1}{m+1}\right\rceil=2$ and $g(m^2+k)=m+1$ .
• If $k=2m$ , then we have $\left\lfloor\dfrac{k}{m}\right\rfloor=\left\lceil\dfrac{k+1}{m+1}\right\rceil=2$ , yielding $f(m^2+k)=m+2$ and $g(m^2+k)=m+1$ .

Thus, $f(n)\neq g(n)$ for all positive integers $n$ which are of the form $m^2+m$ or $m^2+2m$ for some integer $m$ .

For integer $m$ , we have $0<m^2+m<2015$ when $1\leq m\leq 44$ and $0<m^2+2m<2015$ when $1\leq m\leq 43$ .

So, there are $\boxed{87}$ positive integers $n$ satisfied.