15. Given $k$ as a positive integer, the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=3, a_{n+1}=\left(3^{\frac{2}{2 x-1}}-1\right) S_{n}+3\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$

where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$.
Let $b_{n}=\frac{1}{n} \log _{3} a_{1} a_{2} \cdots a_{n}\left(n \in \mathbf{Z}_{+}\right)$, and define
$$
T_{k}=\sum_{i=1}^{2 k}\left|b_{i}-\frac{3}{2}\right| \text {. }
$$

If $T_{k} \in \mathbf{Z}_{+}$, find all possible values of $k$.
15. From the problem, we have
$$
\begin{array}{l}
a_{2}=\left(3^{\frac{2}{2 k-1}}-1\right) \times 3+3=3^{\frac{2 k+1}{2 k-1}} . \\
\text { Also, } a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3, \\
a_{n}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n-1}+3(n \geqslant 2),
\end{array}
$$

Thus, \(a_{n+1}-a_{n}=\left(3^{\frac{2}{2 k-1}}-1\right) a_{n}\)
$$
\begin{array}{l}
\Rightarrow a_{n+1}=3^{\frac{2}{2 k-1}} a_{n} \\
\Rightarrow a_{n}=a_{2}\left(3^{\frac{2}{2 k-1}}\right)^{n-2}=3^{\frac{2 n+2 k-3}{2 k-1}}(n \geqslant 2) .
\end{array}
$$

Clearly, \(n=1\) also fits.
Therefore, the general term formula for the sequence \(\left\{a_{n}\right\}\) is
$$
a_{n}=3^{\frac{2 n+2 k-3}{2 k-1}}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Thus, \(b_{n}=\frac{1}{n} \log _{3} a_{1} a_{2} \cdots a_{n}\)
$$
\begin{array}{l}
=\frac{1}{n} \cdot \frac{1}{2 k-1} \sum_{i=1}^{n}(2 i+2 k-3) \\
=1+\frac{n-1}{2 k-1} .
\end{array}
$$

Therefore, \(b_{n}-\frac{3}{2}=\frac{n-\left(k+\frac{1}{2}\right)}{2 k-1}\). Hence, when \(n \leqslant k\), \(b_{n}-\frac{3}{2} < 0\); when \(n > k\), \(b_{n}-\frac{3}{2} > 0\).
$$
\begin{array}{l}
\text { Then } T_{k}=\sum_{i=1}^{2 k}\left|b_{i}-\frac{3}{2}\right| \\
=\sum_{i=1}^{k}\left(\frac{3}{2}-b_{i}\right)+\sum_{i=k+1}^{2 k}\left(b_{i}-\frac{3}{2}\right) \\
=\frac{k^{2}}{2 k-1} .
\end{array}
$$

Since \(T_{k} \in \mathbf{Z}_{+}\), i.e., \((2 k-1) \mid k^{2}\), we have \((2 k-1) \mid\left(4 k^{2}-1+1\right)\).
Thus, \((2 k-1) \mid 1\).
Hence, \(2 k-1=1 \Rightarrow k=1\).
Therefore, the only possible value of \(k\) is 1.