32. Let the prime factorization of $a$ and $b$ have the powers of $2$ and $5$ as $\alpha_{1}$, $\beta_{1}$ and $\alpha_{2}$, $\beta_{2}$, respectively, then
$$\left\{\begin{array}{l} a \cdot \alpha_{1} + b \cdot \alpha_{2} \geqslant 98 \\ a \cdot \beta_{1} + b \cdot \beta_{2} \geqslant 98 \end{array}\right.$$

and one of (1) and (2) must be an equality.
If (2) is an equality, i.e., $a \cdot \beta_{1} + b \cdot \beta_{2} = 98$, then when $\beta_{1}$ and $\beta_{2}$ are both positive integers, the left side is a multiple of 5. When $\beta_{1}$ or $\beta_{2}$ is zero, the other must be greater than zero, in which case the left side is still a multiple of 5, leading to a contradiction. Therefore, (1) must be an equality.

From $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$, if $\alpha_{1}$ or $\alpha_{2}$ is zero, without loss of generality, let $\alpha_{2} = 0$, then $\alpha_{1} > 0$. In this case, $a \cdot \alpha_{1} = 98$. If $\alpha_{1} \geqslant 2$, then $4 \mid a$, which is a contradiction. Hence, $\alpha_{1} = 1$, and thus $a = 98$. Substituting $a = 98$ into (2), we know $\beta_{1} = 0$, so $b \cdot \beta_{2} > 98$. Combining $\alpha_{2} = 0$, we find that the minimum value of $b$ is 75.

If $\alpha_{1}$ and $\alpha_{2}$ are both positive integers, without loss of generality, let $\alpha_{1} \geqslant \alpha_{2}$. If $\alpha_{2} \geqslant 2$, then $4 \mid a$ and $4 \mid b$, leading to $4 \mid 98$, which is a contradiction. Hence, $\alpha_{2} = 1$. Further, if $\alpha_{1} = 1$, then $a + b = 98$, but $\frac{a}{2}$ and $\frac{b}{2}$ are both odd, so $\frac{a}{2} + \frac{b}{2}$ is even, which is a contradiction. Therefore, $\alpha_{1} > 1$. In this case, if $\beta_{1}$ and $\beta_{2}$ are both positive integers, then $5 \mid a$ and $5 \mid b$, which contradicts $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$. Hence, one of $\beta_{1}$ and $\beta_{2}$ must be zero. If $\beta_{1} = 0$, then from (2) we know $b \cdot \beta_{2} > 98$, in which case the number of trailing zeros in $b^b$ is greater than 98 (since, in this case, $10 \mid b$. When $\beta_{2} = 1$, $b \geqslant 100$, so $10^{100} \mid b^b$. When $\beta_{2} \geqslant 2$, $50 \mid b$, if $b > 50$, then $10^{100} \mid b^b$; if $b = 50$, then $a \cdot \alpha_{1} = 48$, in which case when $\alpha_{1} \geqslant 4$, $2^5 \mid a \cdot \alpha_{1}$, and when $\alpha_{1} \leqslant 3$, $2^4 \nmid a \cdot \alpha_{1}$, both leading to a contradiction, so the number of trailing zeros in $b^b$ is greater than 98).

Similarly, if $\beta_{2} = 0$, then $a \cdot \beta_{1} > 98$, and similarly, the number of trailing zeros in $a^a$ is greater than 98, which is a contradiction. In summary, the minimum value of $ab$ is 7350 (when $(a, b) = (98, 75)$ or $(75, 98)$).