One, (20 points) Given the equation in terms of $x$
$$
x^{2}+2(a+2 b+3) x+\left(a^{2}+4 b^{2}+99\right)=0
$$

has no distinct real roots. How many ordered pairs of positive integers $(a, b)$ satisfy this condition?
Given that $x^{2}+2(a+2 b+3) x+\left(a^{2}+4 b^{2}+99\right)$ $=0$ has no two distinct real roots, therefore,
$$
\Delta=[2(a+2 b+3)]^{2}-4\left(a^{2}+4 b^{2}+99\right) \leqslant 0 \text {. }
$$

Simplifying to $2 a b+3 a+6 b \leqslant 45$, then
$$
(a+3)(2 b+3) \leqslant 54 \text {. }
$$

Since $a, b$ are positive integers, we have
$$
2 b+3 \leqslant \frac{54}{a+3} \leqslant \frac{54}{4} \text {. }
$$

Thus, $2 b+3 \leqslant 13$. Hence, $b \leqslant 5$.
When $b=1$, $a+3 \leqslant \frac{54}{5}$, so $a \leqslant 7$. There are 7 pairs of positive integers that meet the condition.

When $b=2$, $a+3 \leqslant \frac{54}{7}$, so $a \leqslant 4$. There are 4 pairs of positive integers that meet the condition.

When $b=3$, $a+3 \leqslant \frac{54}{9}$, so $a \leqslant 3$. There are 3 pairs of positive integers that meet the condition.

When $b=4$, $a+3 \leqslant \frac{54}{11}$, so $a \leqslant 1$. There is 1 pair of positive integers that meet the condition.

When $b=5$, $a+3 \leqslant \frac{54}{13}$, so $a \leqslant 1$. There is 1 pair of positive integers that meet the condition.

In summary, there are $7+4+3+1+1=16$ pairs of positive integers that meet the condition.