3. Let the pair of positive integers $(x, y)$ be such that $\frac{x^{2}+y^{2}}{11}$ is an integer, and satisfies the condition $\frac{x^{2}+y^{2}}{11} \leqslant 1991$. Find the number of such pairs of positive integers $(x, y)$ (when $a \neq b$, $(a, b)$ and $(b, a)$ are considered different pairs).
3. For any $x \in \mathbf{N}$, if $11 \times x$, then because
$$
\begin{array}{l}
x \equiv \pm 1, \pm 2, \pm 3, \pm 4, \pm 5(\bmod 11), \\
x^{2} \equiv 1,4,9,5,3(\bmod 11),
\end{array}
$$

Therefore, for any $y$, we have $x^{2}+y^{2} \neq 0(\bmod 11)$,
Thus, when $\frac{x^{2}+y^{2}}{11}$ is an integer, it must be that $11 \mid x$. Similarly, $11 \mid y$.
Let $\frac{x^{2}+y^{2}}{11}=11\left(m^{2}+n^{2}\right)$,

Then substituting into the given condition and canceling the factor 11, we get $m^{2}+n^{2} \leqslant 181$.
(*)
Therefore, the number of positive integer pairs $(x, y)$ that satisfy the given condition is the same as the number of positive integer pairs that satisfy condition (1).
Since $n \geqslant 1$, we have $m^{2} \leqslant 181-n^{2} \leqslant 180$.
Since $13^{2}=169<181$,

At this time, there are no pairs $(m, n)$ that satisfy condition (*).
(3) When one of $m, n$ takes one of $10,11,12,13$, because
$$
\begin{array}{l}
9^{2} \leqslant 181-10^{2}=81<10^{2}, \\
7^{2} \leqslant 181-11^{2}=60<8^{2}, \\
6^{2} \leqslant 181-12^{2}=37<7^{2}, \\
3^{2} \leqslant 181-13^{2}=12<4^{2},
\end{array}
$$

Therefore, the number of pairs $(m, n)$ that satisfy condition (*) is $2 \times(9+7+6+3)=50$ (pairs).
In summary, the number of positive integer pairs $(x, y)$ that satisfy the given condition is $81+50=131$ pairs.