Solution:

`13` pairs of positive integers `a` and `b`

Step by step solution:

$\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{2001}$

$\Leftrightarrow 2001a + 2001b = ab\qquad (*)$

$\Leftrightarrow ab - 2001a - 2001b + 2001^2 = 2001^2$

$\Leftrightarrow (a-2001)(b-2001) = 2001^2\qquad (**)$

From $(*)$ we have:

$a = \dfrac{2001b}{b - 2001}$

$\Leftrightarrow b - a = \dfrac{b(b-4002)}{b-2001}$

We also have: $b> a$ and $a;\,b$ is a positive integer

So that $\begin{cases}b- a >0\\a>0\\b >0\end{cases}$

$\Leftrightarrow \begin{cases}\dfrac{b-4002}{b-2001} >0\\\dfrac{2001b}{b-2001}>0\\b >0\end{cases}$

$\Leftrightarrow b - 4002 >0$

$\Leftrightarrow b > 4002$

$\Leftrightarrow b - 2001 > 2001$

In the other side, from $(**)$ we have:

$\qquad \qquad 2001^2 = \begin{cases}1578.2523\\841.4761\\667.6003\\529.7569\\261.15341\\207.19343\\87.46023\\69.58029\\29.13069\\23.174087\\9.444889\\3.1334667\\1.4004001\end{cases}$

$\Leftrightarrow (a-2001)(b-2001) = \begin{cases}1578.2523\\841.4761\\667.6003\\529.7569\\261.15341\\207.19343\\87.46023\\69.58029\\29.13069\\23.174087\\9.444889\\3.1334667\\1.4004001\end{cases}$

We have a list of value as below:

$\begin{array}{|l|r|}
\hline
a -2001&1587&841\,&667\,&529\,\,&261\,\,\,&207\,\,\,&87\,\,\,\,&69\,\,\,\,&29\,\,\,\,&23\quad &9\quad\,\,&3\quad\,\,&1\quad\,\,\\
\hline
b-2001&2523&4761&6004&7569&15341&19343&46023&58026&13069&174087&444889&1334667&4004001\\
\hline
\quad \,\,\,a&3588&2842&2668&2530&2262\,&2208\,&2088\,&2070\,\,&2030\,\,&2024\,\,\,&2010\,\,\,&2004\,\,\,\,\,&2002\,\,\,\,\\
\hline
\quad\,\,\,b&4524&6762&8004&9570&17342&21344&48024&60030&140070&176088&446890&1336668&4006002\\
\hline
\end{array}$

Therefore we get `13` pairs of positive integers a and b which meet the problem requirements