10. If $\frac{1^{2}+3^{2}+\cdots+(2 n-1)^{2}}{2^{2}+4^{2}+\cdots+(2 n)^{2}}=\frac{13}{14}$, then
$$
n=
$$
10.20.

From $\sum_{k=1}^{2 n} k^{2}=\frac{27}{14} \sum_{k=1}^{n}(2 k)^{2}$, we get
$$
\begin{array}{l}
\frac{2 n(2 n+1)(4 n+1)}{6} \\
=\frac{27}{14} \times 4 \times \frac{n(n+1)(2 n+1)}{6},
\end{array}
$$

which means $4 n+1=\frac{27}{7}(n+1)$.
Therefore, $n=20$.