10. For real numbers $x_{1}, x_{2}, \cdots, x_{19}$, the maximum integer not exceeding
$$
f\left(x_{1}, x_{2}, \cdots, x_{19}\right)=\sum_{k_{1}=0}^{1} \sum_{k_{2}=0}^{1} \cdots \sum_{k_{19}=0}^{1}\left|k_{1} x_{1}+k_{2} x_{2}+\cdots+k_{19} x_{19}-1\right|
$$

is $\qquad$ .
Consider the general case. We divide the set $A=\{0,1\}^{n}$ into: $A=A_{0} \cup A_{1} \cup \cdots \cup A_{n}$,
where $A_{k}=\left\{\beta=\left(k_{1}, k_{2}, \cdots, k_{n}\right) \in A \mid k_{1}+k_{2}+\cdots+k_{n}=k\right\}, k=0,1, \cdots, n$.
Obviously, the number of elements in $A_{k}$ is $\left|A_{k}\right|=C_{n}^{k}$, and let $X=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$, then
$$
f(X)=\sum_{\beta \in A}|\beta \cdot X-1|=\sum_{k=0}^{n} \sum_{\beta \in A_{k}}|\beta \cdot X-1| \geqslant \sum_{k=0}^{n}\left|\sum_{\beta \in A_{k}}(\beta \cdot X-1)\right| \text {. }
$$

Since $B_{k}=\sum_{\beta \in A_{k}}(\beta \cdot X-1)=\frac{k}{n} C_{n}^{k}\left(x_{1}+x_{2}+\cdots+x_{n}\right)-C_{n}^{k}$
$=C_{n-1}^{k-1}\left(x_{1}+x_{2}+\cdots+x_{n}\right)-C_{n}^{k}$, and $C_{n-1}^{k-1}=C_{n-1}^{n-k}=C_{n-1}^{n+1-k-1}$, thus
$$
B_{k}-B_{n+1-k}=C_{n}^{n+1-k}-C_{n}^{k}=C_{n}^{k-1}-C_{n}^{k} \Rightarrow\left|B_{k}\right|+\left|B_{n+1-k}\right| \geqslant C_{n}^{k}-C_{n}^{k-1} \text {. }
$$

Let $B_{n+1}=0$, so
$$
f(X) \geqslant \sum_{k=0}^{n+1}\left|B_{k}\right| \geqslant \sum_{k=0}^{\left[\frac{n}{2}\right]}\left(\left|B_{k}\right|+\left|B_{n+1-k}\right|\right) \geqslant \sum_{k=0}^{\left[\frac{n}{2}\right]}\left(C_{n}^{k}-C_{n}^{k-1}\right)=C_{n}^{\left[\frac{n}{2}\right]} .
$$

When $n=2 m$ or $n=2 m-1$, taking $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{m}$ can make the above equality hold (at this time $B_{m}=0$):
$$
f(X)=B_{0}+B_{1}+\cdots+B_{m-1}-B_{m+1}-\cdots-B_{n}=\sum_{k=0}^{n-m}\left(B_{k}-B_{n+1-k}\right)=C_{n}^{n-m} .
$$

In summary, the minimum value of $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is $C_{n}^{\left[\frac{n}{2}\right]}$, and the result of this problem is $C_{19}^{10}=92378$.