15. For any real number $x$, let $\lceil x\rceil$ denote the smallest integer that is greater than or equal to $x$ and $\lfloor x\rfloor$ denote the largest integer that is less than or equal to $x$ (for example, $\lceil 1.23\rceil=2$ and $\lfloor 1.23\rfloor=1$ ). Find the value of
$$
\sum_{k=1}^{2010}\left[\frac{2010}{k}-\left\lfloor\frac{2010}{k}\right\rfloor\right] \text {. }
$$
15. Answer: 1994
Consider $k=1,2, \ldots, 2010$. If $k \mid 2010$, then $x:=\frac{2010}{k}-\left\lfloor\frac{2010}{k}\right\rfloor=0$, so $\lceil x\rceil=0$. If
$k \backslash 2010$, then $0<y:=\frac{2010}{k}-\left\lfloor\frac{2010}{k}\right\rfloor<1$, so $\lceil y\rceil=1$.
Since the prime factorization of 2010 is $2 \times 3 \times 5 \times 67$, we see that 2010 has 16 distinct divisors. Hence
$$
\begin{aligned}
\sum_{k=1}^{2010}\left\lceil\frac{2010}{k}-\left\lfloor\frac{2010}{k}\right\rfloor\right\rceil & =\sum_{k: 2010}\lceil x\rceil+\sum_{k, 2010}\lceil y\rceil \\
& =\text { number of non-divisor of } 2010 \text { among } k \\
& =2010-16=1994
\end{aligned}
$$