Let $A ∈ M_n$ be positive and let $x = [x_i ]$ be the Perron vector of $A$.
Suppose $\rho (A) = \mathop {\min }\limits_i \sum\limits_{j = 1}^n {{a_{ij}}} $ or $\rho (A) = \mathop {\max }\limits_i \sum\limits_{j = 1}^n {{a_{ij}}} $
Why does $x_1 = · · · = x_n$?
(Note : $\rho (A) = \max \{ \left| \lambda \right|:\lambda $ is eigenvalue of $A$ $\}$).
Suppose the eigenvector equation $Ax=\lambda x$ for the largest eigenvalue. This means $$\sum_{j=1}^N a_{ij}x_j=\lambda x_i$$ for every $i$. If we also know that $$\sum_{j=1}^N a_{ij}=\lambda,$$ then $x_1=\cdots=x_N$ is obviously a solution.
These are $N$ linear equations with $N$ unknowns, so if the equations are linearly independent, the system has a single solution and this is the end. If they are not L.I., you could have other solutions and then $\lambda$ would be degenerate.
Maybe you should look up the Perron-Frobenius theorem.
EDIT.
The above part of the answer assumes $A$ is stochastic, i.e. every row sum is equal to $\lambda$. But this can be proved from the condition $\lambda={\rm max}\sum_{j}a_{ij}$, as follows.
Suppose the first row has the largest sum, $\sum_{j}a_{1j}=\lambda$ and suppose $x_1=1$ so $\sum_{j}a_{1j}x_j=\lambda$ as well.
If $x$ is constant, stochasticity of $A$ is obvious. So assume $x$ is not constant. It is not possible for $x_i<1$ for all $i>2$ or this would contradict the above. So some $x_i$ must be larger than $1$. Suppose the largest $x_i$ is $x_2$.
Then $\sum_{j}a_{2j}x_j=\lambda x_2.$ On the other hand, $\sum_{j}a_{2j}x_j<(\sum_{j}a_{2j}) x_2.$ It follows that $\lambda<\sum_{j}a_{2j}$.
But this contradicts the assumption that the first row had the largest sum.
Therefore, all rows must have the same sum.