22. Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}(n \geqslant 2)$, and satisfy $\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1}=1$. Try to find $\operatorname{maxmmax}_{1 \leqslant k \leqslant n}\left|x_{k}\right|$.
22. First, for a given $k$, find the maximum value of $\left|x_{k}\right|$.

In fact, the given equation is $x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}+\cdots+\left(x_{n-1}+x_{n}\right)^{2}+x_{n}^{2}=2$.
Notice the identity $\left|x_{k}\right|=\left|x_{1}-\left(x_{1}+x_{2}\right)+\cdots+(-1)^{k-1}\left(x_{k-1}+x_{k}\right)\right|$,
$$
\left|x_{k}\right|=\left|x_{n}-\left(x_{n-1}+x_{n}\right)+\cdots+(-1)^{n-k}\left(x_{k}+x_{k-1}\right)\right| \text {. }
$$

By the Cauchy-Schwarz inequality, we have $x_{k}^{2} \leqslant k\left[x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\cdots+\left(x_{k-1}+x_{k}\right)^{2}\right]$,
$$
x_{k}^{2} \leqslant(n-k+1)\left[x_{n}^{2}+\left(x_{n-1}+x_{n}\right)^{2}+\cdots+\left(x_{k}+x_{k-1}\right)^{2}\right] \text {. }
$$

From equations (4) and (5), combined with equation (1), we get $\frac{x_{k}^{2}}{k}+\frac{x_{k}^{2}}{n-k+1} \leqslant 2$.
Solving this, we get $\left|x_{k}\right| \leqslant \sqrt{\frac{2 k(n+1-k)}{n+1}}(k=1,2, \cdots, n)$.
By the condition for equality in the Cauchy-Schwarz inequality, we know that equality in equation (6) holds if and only if equalities in equations (4) and (5) hold simultaneously, i.e.,
$$
\begin{array}{c}
x_{1}=-\left(x_{1}+x_{2}\right)=x_{2}+x_{3}=\cdots=(-1)^{k-1}\left(x_{k-1}+x_{k}\right), \\
x_{k}+x_{k+1}=-\left(x_{k-1}+x_{k-2}\right)=\cdots=(-1)^{n-k} x_{n} .
\end{array}
$$

From equation (7), for $1<i \leqslant k$, we have $x_{1}=(-1)^{i-1}\left(x_{i-1}+x_{i}\right)$, i.e., $x_{i}+x_{i-1}=(-1)^{i-1} x_{1}$. Then $(-1)^{i} x_{i}-(-1)^{i-1} x_{i-1}=-x_{1}$. Hence $(-1)^{i} x_{i}=-x_{1}+\left(-x_{1}\right)(i-1)=-x_{1} i$. Therefore, $x_{i}=(-1)^{i-1} i x_{1}(i=2,3, \cdots, k)$. Let $i=k$, we get $x_{1}=\frac{x_{k}}{k}(-1)^{k-1}$. Therefore, $x_{i}=x_{k}(-1)^{-k} \cdot \frac{i}{k}(i=1,2, \cdots, k-1)$.
Similarly, from equation (8), we can find $x_{j}=x_{k}(-1)^{j-k} \cdot \frac{n+1-j}{n-k+1}(j=k+1, \cdots, n)$.
The above shows that there exists a set of real numbers $x_{1}, x_{2}, \cdots, x_{n}$ that satisfy the given conditions and make equation (6) hold with equality.
Therefore, for a given $k$, we have $\max \left|x_{k}\right|=\sqrt{\frac{2 k(n+1-k)}{n+1}}(1 \leqslant k \leqslant n)$.
As $k$ varies from 1 to $n$, it is not difficult to prove that $k(n+1-k) \leqslant\left[\frac{n+1}{2}\right]\left[\frac{n+2}{2}\right]$.
Equality holds if and only if $k=\left[\frac{n+1}{2}\right]$.
Thus, $\max _{1 \leqslant n} \max \left|x_{k}\right|=\sqrt{\frac{2}{n+1}\left[\frac{n+1}{2}\right]\left[\frac{n+2}{2}\right]}$, where $[x]$ is the floor function.