27. $n$ is a positive integer, $X_{\mathrm{n}}$ represents the set of all $n$-tuples of non-negative integers $\left(x_{1}, x_{2}, \cdots, x_{\mathrm{n}}\right)$, where $\left(x_{1}, x_{2}, \cdots, x_{\mathrm{n}}\right)$ satisfies the equation $x_{1}+x_{2}+\cdots+x_{\mathrm{n}}$ $=n$. And $Y_{\mathrm{n}}$ represents the set of all $n$-tuples of non-negative integers $\left(y_{1}, y_{2}, \cdots, y_{\mathrm{n}}\right)$, where $\left(y_{1}, y_{2}, \cdots, y_{\mathrm{n}}\right)$ satisfies the equation $y_{1}+y_{2}+\cdots+y_{\mathrm{n}}=2 n$. If for $1 \leqslant i$ $\leqslant n$, $x_{i} \leqslant y_{i}$, then the number of compatible pairs of $n$-tuples from $X_{\mathrm{n}}$ and $Y_{\mathrm{n}}$ is a perfect square.
27. Consider a special set $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ in $X_{n}$. Suppose it is compatible with $\left(y_{1}, y_{2}, \cdots, y_{n}\right)$ in $Y_{n}$. For $1 \leqslant i \leqslant n$, let $z_{1}=y_{1}-x_{1} \geqslant 0$, because
$$
\boldsymbol{z}_{1}+\boldsymbol{z}_{2}+\cdots+\boldsymbol{z}_{n}=2 n-n=n,
$$

so, $\left(z_{1}, z_{2}, \cdots, z_{n}\right) \in X_{n}$.
Conversely, for any $\left(z_{1}, z_{2}, \cdots, z_{n}\right)$ in $X_{n}$, let $y_{1}=x_{1}+z_{1} \geqslant x_{1}, 1 \leqslant i \leqslant n$. Because
$$
y_{1}+y_{2}+\cdots+y_{n}=n+n=2 n,
$$

so, $\left(y_{1}, y_{2}, \cdots, y_{n}\right) \in Y_{n}$.
Thus, it is compatible with $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$. This implies that there are indeed $\left|X_{n}\right|$ $n$-tuples in $Y_{n}$ that are compatible with $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$. Since $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ can be any $n$-tuple in $X_{n}$, the total number of compatible pairs is $\left|X_{n}\right|^{2}$.