Is it true that for any set of $n$ distinct positive real numbers $$\lambda_1 < \cdots < \lambda_n$$ there is a real symmetric matrix $A_{n \times n}$ with eigenvalues $\lambda_1, \ldots, \lambda_n$, and $A>0$ entrywise?
If so, is there an algorithm to find the $A$ for any given input?
Here's a sketch of a proof:
First, you can put the eigenvalues on the diagonal to get a "solution" $A_0$ that doesn't quite satisfy $A > 0$. Let $F: Sym \to R^n$ be the map that produces the $n$ eigenvalues in increasing order. (Here "Sym" denotes symmetric n x n matrices).
Note that $dF/da_{ii}(A_0) = e_i$.
Now change $A_0$ to $A_1$ by adding a very small number $c$ to every off-diagonal entry. We can assume that at $A_1$, we have $$ dF/da_{ii}(A_1) \approx e_i. $$ In particular, for small enough $c$, these $n$ partial derivatives are linearly independent.
Now do a gradient search: the problem is that $F(A_1)$ is not the list of eigenvalues that you want, but by altering the diagonal terms, you can (by the observation about the partials) arrange to get closer to the eigenvalues that you wanted. In particular, for $c$ small enough, the alteration you need to make will be smaller than the smallest of the eigenvalues, hence all entries of the resulting matrix will still be be positive.