It's an exercise in, e.g., Bredon's "Geometry and Topology" that a square matrix with positive real entries necessarily has a positive real eigenvalue. (Sketch: consider the intersection of the positive orthant with the closed unit sphere in $\mathbb R^n$; this set is homeomorphic to the closed disk $D^{n-1}$; if the matrix is $A$, consider the map $D^{n-1} \rightarrow D^{n-1}$ defined by $x \mapsto \frac{A x}{\|A x\|}$. Brouwer's fixed point theorem says that this map has a fixed point, and thus that $A$ has a positive real eigenvalue.)
I'm wondering whether there is an elementary proof of this fact, e.g., using only techniques of linear algebra.
The Rayleigh-Ritz method for finding the largest eigenvalue for a real symmetric matrix $A$ is to maximize $(Ax,x)$ subject to the constraint $\|x\|^{2}=1$. Such a maximum exists because $(Ax,x)=\sum_{j}\sum_{k}a_{j,k}x_{j}x_{k}$ is a continuous function of $(x_{1},x_{2},\cdots,x_{N})$, and the unit sphere is closed and bounded. If $\lambda$ is that maximum value, and if $(Ax_{0},x_{0})=\lambda$, then one can show that $Ax_{0}=\lambda x_{0}$ automatically follows from the fact that $f(x)=((\lambda I-A)x,x)$ has an absolute minimum at $x=x_{0}$. A proof is accomplished by noting that $h(t)=f(x_{0}+ty)$ satisfies $h'(0)=0$ for all $y$, from which it follows that $((\lambda I-A)x_{0},y)=0$ for all $y$.
For your proof: Suppose that $A$ has all positive entries. It isn't hard to see that if $x=(x_{1},x_{2},\cdots,x_{N})$ is of unit length, then $y=(|x_{1}|,|x_{2}|,\cdots,|x_{N}|)$ has unit length and $(Ax,x) \le (Ay,y)$. It follows from the above discussion that there is an eigenvector corresponding to the maximum eigenvalue with all positive entries.
From my understanding of the situation, some Analysis is required. You may assume a result that uses Analysis, or you may use it directly such as through the Ritz Method. It may be overt, or it may be hidden. But somewhere it pops up, for the same reason that there is no purely algebraic proof of the fundamental theorem of Algebra.