One, (20 points) Find the positive integer solutions of the indeterminate equation
$$
29 a+30 b+31 c=2196
$$
One, transform the original equation into
$$
\left\{\begin{array}{l}
29(a+b+c)+(b+2 c)=2196 \\
31(a+b+c)-(2 a+b)=2196 .
\end{array}\right.
$$

Since \(a, b, c\) are positive integers, from equation (1) we get
$$
\begin{array}{l}
29(a+b+c)=2196-(b+2 c) \\
\leqslant 2196-(1+2 \times 1)=2193 .
\end{array}
$$

Therefore, \(a+b+c \leqslant 75 \frac{18}{29}\).
From equation (2) we get
$$
\begin{array}{l}
31(a+b+c)=2196+(2 a+b) \\
\geqslant 2196+(2 \times 1+1)=2199 .
\end{array}
$$

Therefore, \(a+b+c \geqslant 70 \frac{29}{31}\).
From equations (3) and (4) we get
$$
a+b+c=71,72,73,74 \text { or } 75 \text {. }
$$

When \(a+b+c=71\), combining with the original equation, we solve to get
$$
b=5-2 a, c=a+66 \text {. }
$$

From \(b \geqslant 1\), we get \(5-2 a \geqslant 1\). Solving this, we get \(a \leqslant 2\), so \(a=1\) or 2. In this case, the original equation has 2 sets of positive integer solutions.
Similarly, when \(a+b+c=72,73,74,75\), we can respectively find that the original equation has \(17, 33, 24, 10\) sets of positive integer solutions.
Therefore, the total number of positive integer solutions to the original equation is
$$
2+17+33+24+10=86 \text { (sets). }
$$