Consider the nondecreasing vector $v=(1,1,\ldots,1)$. It's not too hard to show that two nondecreasing nonzero vectors orthogonal to $v$ must have a positive inner product. Here is a proof.

Let $a=(a_1,\ldots, a_n)$ and $b=(b_1,\ldots, b_n)$ be orthogonal to $v$. Suppose that $a_j$ is the last negative entry of $a$ and $b_k$ is the last negative entry of $b$, for $k>j$. Then

\begin{equation}\sum_{i=1}^j |a_ib_i| \ge \sum_{i=1}^j |a_i b_{j+1}| \ge \sum_{i=j+1}^k |a_ib_{j+1}|\ge \sum_{i=j+1}^k |a_ib_i|\end{equation}

But the sum defining the inner product is

\begin{equation}\sum_{i=1}^j |a_ib_i| - \sum_{i=j+1}^k |a_ib_i| + \sum_{i=k+1}^n |a_ib_i|\end{equation}

So this is nonzero as long as $a_n$ and $b_n$ are nonzero; since $a$ and $b$ are orthogonal to $v$, $a_n$ is zero if and only if $a$ is zero, and likewise for $b.$

Now, a generic nondecreasing vector is of the form $av+u$ for $u$ nondecreasing and orthogonal to $v$.

Assume we have three mutually orthogonal nonzero nondecreasing vectors $v_i= a_i v +u_i$ for $i\in\{1,2,3\}$. No more than one $u_i$ can be zero; suppose $u_1$ and $u_2$ are nonzero. Then $a_1$ and $a_2$ must have opposite signs for $v_1$ and $v_2$ to be orthogonal (in particular, neither can be zero). Now $u_3$ cannot be zero because then $v_3$ would not be orthogonal to $v_1$ or $v_2$. Then by the same argument, $a_3$ must have a sign opposite to both $a_1$ and $a_2$, a contradiction.