26 Determine all non-empty subsets $A, B, C$ of $\mathbf{N}^{*}$ such that:
(1) $A \cap B=B \cap C=C \cap A=\varnothing$;
(2) $A \cup B \cup C=\mathbf{N}^{*}$;
(3) For all $a \in A, b \in B, c \in C$, we have $a+c \in A, b+c \in B, a+b \in C$
Let $t$ be the smallest element in $C$, then $1,2, \cdots, t-1 \in A \cup B$. By the condition $a+t \in A, b+t \in B$, we know that all positive integers not divisible by $t$ are in $A \cup B$, hence $C$ contains only multiples of $t$, so for $\forall a, b, t \mid a+b$.

If $t \geqslant 4,1,2,3$ at least two of these numbers belong to $A$ or $B$. Without loss of generality, assume two of these numbers $a_{1}, a_{2} \in A$, take any $b \in B$, then $t\left|a_{1}+b, t\right| a_{2}+b$ cannot both be true.
If $t=1$, without loss of generality, assume $a<b$, then by $b=a+(b-a) \cdot 1$ we know $b \in A$, a contradiction!
If $t=2$, without loss of generality, assume $1 \in A$, then all odd numbers belong to $A$, but by $2 \mid 1+b$ we know $b$ is odd, a contradiction! Therefore, $t=3$.

Without loss of generality, assume $1 \in A$, then $3 k+1 \in A$, take $b$ such that $3 \mid b+1$, hence $b \equiv 2(\bmod 3)$, this way $2 \in B$, so $(3 k+1)+2 \in C$, i.e., $C$ is exactly all multiples of 3.
At this point
$$
\begin{array}{l}
(A, B, C)=\left(\left\{3 k-2 \mid k \in \mathbf{Z}^{+}\right\},\right. \\
\left.\left\{3 k-1 \mid k \in \mathbf{Z}^{+}\right\},\left\{3 k \mid k \in \mathbf{Z}^{+}\right\}\right),
\end{array}
$$

Therefore, the required $(A, B, C)$ is
$$
\left(\left\{3 k-2 \mid k \in \mathbf{Z}^{+}\right\},\left\{3 k-1 \mid k \in \mathbf{Z}^{+}\right\},\left\{3 k \mid k \in \mathbf{Z}^{+}\right\}\right)
$$

or $\left(\left\{3 k-1 \mid k \in \mathbf{Z}^{+}\right\},\left\{3 k-2 \mid k \in \mathbf{Z}^{+}\right\},\left\{3 k \mid k \in \mathbf{Z}^{+}\right\}\right)$.