Let $L = L^{\sf T} \in \mathbb{R}^{n\times n}$ be a (weighted) Laplacian matrix of a connected undirected graph. For those not familiar with Laplacians (they are positive semidefinite); for simplicity just take $L = n I_n - 1_{n \times n}$, where $1_{n \times n}$ is the matrix of all ones. I have a vector valued function $f(x) = (f_1(x_1),\ldots,f_n(x_n))$, with whatever regularity you desire. Note the component-by-component decoupling in $f$.
Under what conditions on $f$ does it hold that $$ x^{\sf T}Lf(x) \geq 0\,,\qquad \qquad \forall x \in \mathbb{R}^n? $$
Obviously if $f(x) = x$, it holds. Another simple case I can identify is when the components of $f(x)$ are all the same function: $f_i = f_j = h$, and $h$ is a monotone function. Then the product reduces to
\begin{align*} x^{\sf T}Lf(x) &= \sum_{i,j=1}^n {(f_i(x_i)-f_i(x_j))(x_i-x_j)}\\ &= \sum_{i,j=1}^n \underbrace{(h(x_i)-h(x_j))(x_i-x_j)}_{\geq 0 \forall x_i, x_j}\\ & \geq 0 \end{align*}
Beyond that, I'm looking for some conditions on $f_1,\ldots,f_n$, possibly accompanied jointly with conditions on $L$.
Assuming the $f_i$ are continuous and $n>1$, your example covers all solutions.
Let $I=\{1,\ldots,n\}$ and let $E\subseteq I\times I$ be the set of edges (including each edge twice for the two directions). Fix $a\in\mathbb R$ and let $m=\min_{i\in I}f_i(a)$. Let $$\begin{eqnarray*} J=\{i\in I\mid f_i(a)=m\},\\ K=\{i\in I\mid f_i(a)>m\}. \end{eqnarray*}$$ Note that $J\neq\emptyset$ and $I=J\sqcup K$. Pick any $b<a$ and define the vector $x$ by $x_i=a$ for $i\in J$, $x_i=b$ for $i\in K$. Then $$\begin{eqnarray*} 0&\leq&x^TLf(x)\\ &=&\frac12\sum_{(i,j)\in E}(f_i(x_i)-f_j(x_j))(x_i-x_j)\\ &=&\sum_{(i,j)\in E\cap(J\times K)}(f_i(a)-f_j(b))(a-b) \end{eqnarray*}$$ Hence $$ 0\leq\sum_{(i,j)\in E\cap(J\times K)}(f_i(a)-f_j(b)). $$ Taking the limit as $b\rightarrow0$, $$ 0\leq\sum_{(i,j)\in E\cap(J\times K)}(f_i(a)-f_j(a)). $$ However $f_i(a)-f_j(a)<0$ for any $(i,j)\in J\times K$. Thus $E\cap(J\times K)=\emptyset$. Since the graph was assumed connected, $K=\emptyset$. Thus $f_i(a)=m$ for all $i$. Since $a$ was arbitrary, all the $f_i$ must equal the same function $h$.
Now (assuming $n>1$), break $I$ into two disjoint nonempty sets $J\sqcup K$. Suppose $b<a$. The same calculation as above gives $$ 0\leq|E\cap(J\times K)|(h(a)-h(b)). $$ Since the graph is connected, $E\cap(J\times K)\neq\emptyset$, so $h(b)\leq h(a)$. Hence $h$ is nondecreasing.