One, (20 points) Given that $a$ and $b$ are positive integers. Find the smallest positive integer value that
$$
M=3 a^{2}-a b^{2}-2 b-4
$$

can take.
Given that $a$ and $b$ are positive integers, to make
$$
M=3 a^{2}-a b^{2}-2 b-4
$$

a positive integer, it is evident that $a \geqslant 2$.
When $a=2$, $b$ can only be 1, in which case, $M=4$.
When $a=3$, $b$ can only be 1 or 2.
If $b=1$, then $M=18$; if $b=2$, then $M=7$.
When $a=4$, $b$ can only be 1, 2, or 3.
If $b=1$, then $M=38$; if $b=2$, then $M=24$;
if $b=3$, then $M=2$.
If $M=1$, that is,
$$
3 a^{2}-a b^{2}=2 b+5 \text {, }
$$

and since $2 b+5$ is an odd number, $a$ must be odd and $b$ must be even. In this case, $3 a^{2}-a b^{2}$ leaves a remainder of 3 when divided by 4, while $2 b+5$ leaves a remainder of 1 when divided by 4, so equation (1) cannot hold.
Thus, $M \neq 1$.
Therefore, the smallest positive integer value that $M=3 a^{2}-a b^{2}-2 b-4$ can take is 2.