All four traits

To generate the minimum number with all four traits: 5 of them do not have white paws. Take the 8 that do not have green eyes from the 95 that do have white paws. That leaves 87 with both. Take the 12 that do not have black bellies out of this 87, leaving a minimum of 75 with all three traits. Then take the 15 that do not have curly tails out of this 75, leaving a minimum of 60 with all four characteristics.

Let $W,G,B,C$ be the sets of white-pawed, green-eyed, black-bellied and curly-tails kittens. Then $\begin{aligned} 100 \; \ge \; |W \cup G| & = \; |W| + |G| - |W \cap G| \; = \; 187 - |W \cap G| \\ 100 \; \ge \; |B \cup C| & = \; |B| + |C| - |B \cap C| \; = \; 173 - |B \cap C| \end{aligned}$ so that $|W \cap G| \ge 87$ and $|B \cap C| \ge 73$ , with equality exactly when $W \cup G = B \cup C = \mathcal{C}$ , the set of all $100$ kittens. But then $100 \; \ge \; \big|(W \cap G) \cup (B \cup C)\big| \; = \; |W \cap G| + |B \cap C| - |W \cap G \cap C \cap B| \; \ge \; 160 - |W \cap G \cap B \cap C|$ so that $|W \cap G \cap B \cap C| \ge 60$ .

Moreover we achieve $|W \cap G \cap B \cap C| = 60$ provided that $W \cup G = B \cup C = (W \cap G) \cup (B \cap C) = \mathcal{C}$ .This is possible, if we have $\begin{array}{rclcrcl} |W' \cap G \cap B \cap C| & = & 5 & \hspace{2cm} & |W \cap G' \cap B \cap C| & = & 8 \\ |W \cap G \cap B' \cap C| & = & 12 & & |W \cap G \cap B \cap C'| & = & 15 \\ |W \cap G \cap B \cap C| & = & 60 \end{array}$ Thus the minimum number of kittens with all four traits is $\boxed{60}$ .

Just add up 95+92+88+85=360. To minimize the number of cats that have four traits, we should make every cat have at least 3 traits. $3*100=300$ , so there are at least $360-300=60$ cats who must have 4 traits. Therefore, the answer is $\fbox{60}$