3. Given $101 \mid a$, and
$$
a=10^{j}-10^{i}(0 \leqslant i<j \leqslant 99) \text {. }
$$

Then the number of $a$ that satisfy the condition is. $\qquad$ .
3.1200.

From the problem, we know
$$
\begin{array}{l}
a=10^{j}-10^{i} \\
=10^{i}\left(10^{j-i}-1\right)(0 \leqslant i<j \leqslant 99) .
\end{array}
$$

Obviously, $101 \mid \left(10^{j-i}-1\right)$.
Notice that, $10^{2} \equiv-1(\bmod 101)$,
$$
10^{4} \equiv 1(\bmod 101) \text {. }
$$

Thus, the order of 10 modulo 101 is 4.
Therefore, $4 \mid (j-i)$.
In the set $0,1, \cdots, 99$, the numbers that are congruent to $0, 1, 2, 3$ modulo 4 each have 25 numbers, hence, there are $4 \mathrm{C}_{25}^{2}=1200$ such $a$ values.