Is this type of matrix always positive semidefinite?

Yes, it is positive semidefinite. This can be easily proved by mathematical induction on $n$.

To stress its size as well as its dependence on the parameters $b_i$s, let us denote the matrix generated by $(b_1,\ldots,b_n)\in[0,1]^n$ as described in your question as $A(b_1,\ldots,b_n)$. In the inductive step, let $b_i=\min(b_1,\ldots,b_n)$. Then \begin{aligned} A(b_1,\ldots,b_n) &=\left[\begin{array}{c|c|c} A(b_1,\,\ldots,\,b_{i-1}) &\begin{matrix}b_i\\ \vdots\\ b_i\end{matrix} &\begin{matrix}b_i&\cdots&b_i\\ \vdots&&\vdots\\ b_i&\cdots&b_i\end{matrix}\\ \hline \begin{matrix}b_i&\cdots&b_i\end{matrix}&b_i&\begin{matrix}b_i&\cdots&b_i\end{matrix}\\ \hline \begin{matrix}b_i&\cdots&b_i\\ \vdots&&\vdots\\ b_i&\cdots&b_i\end{matrix} &\begin{matrix}b_i\\ \vdots\\ b_i\end{matrix} &A(b_{i+1},\,\ldots,\,b_n) \end{array}\right]\\ &=\left[\begin{array}{c|c|c} A(b_1-b_i,\,\ldots,b_{i-1}-b_i)&0&0\\ \hline 0&0&0\\ \hline 0&0&A(b_{i+1}-b_i,\,\ldots,\,b_n-b_i) \end{array}\right]+b_iE \end{aligned} where $E$ denotes the all-one matrix. Since $b_k-b_i\in[0,1]$ for every $k$, we see that $A(b_1,\ldots,b_n)$ is positive semidefinite, by induction assumption.