Solution: Obviously, when $T \geqslant 2010$,
$$
\frac{2010}{T} \leqslant 1, \frac{2010+T}{2 T} \leqslant 1,
$$

which does not meet the requirement.
Therefore, $T < 2010$ and
$$
\begin{array}{l}
\frac{2010}{T}>\frac{2010+T}{2 T} .
\end{array}
$$

From this, we know
$$
\begin{array}{l}
45,
$$
meaning that there are at most four positive integers, 6, 7, 8, and 9, between $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$, which is a contradiction.

In summary, $T$ is a positive integer such that $168 \leqslant T \leqslant 200$, totaling 33.
Initial 285 Question: How many positive integers $T$ are there such that there are exactly 5 different positive integers between $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$ (not including $\frac{2010}{T}$ and $\left.\frac{2010+T}{2 T}\right)$?