I am making some mistake computing an example with root systems, but I have so far been unable to find it. Let $J$ be the $6$ by $6$ matrix with $1$s on the antidiagonal, and take $G = \textrm{SO}_6(\mathbb{C})$ as those $g \in \textrm{SL}_6(\mathbb{C})$ for which $^tg Jg = J$.
The standard maximal torus $T$ for $G$ is the elements of the form $\textrm{diag}(t_1,t_2,t_3,t_3^{-1},t_2^{-1},t_1^{-1})$. If $e_1, e_2, e_3$ is the standard basis for $X(T)$, then a system of positive roots of $T$ in $G$ is
$$\Phi^+ = \{e_1 - e_2, e_1 - e_3, e_1 + e_3,e_1 + e_2, e_2 - e_3, e_2 + e_3\}$$
for which the base is $\alpha_1 = e_1-e_2, \alpha_2 = e_2 - e_3, \alpha_3 = e_2 + e_3$. The Weyl group $W = N_G(T)/T$ can be identified with the group of signed permutations of $1, 2, 3$ where you are allowed to change exactly two signs at a time. Let $s_i$ be the simple reflection corresponding to $\alpha_i$. Then $s_1 = (1 \, 2), s_2 = (2 \, 3)$, and $s_3$ swaps $2$ and $-2$, as well as $3$ and $-3$.
Consider the Weyl group element
$$w_0 = s_1s_3s_1 \in W$$
It swaps $1$ and $-1$, and swaps $3$ and $-3$, and leaves $2$ and $-2$ alone. The length of $w_0$ should be the number of positive roots it makes negative. It seems like it makes four positive roots negative: $e_1 - e_2, e_1 - e_3, e_1 +e_3,$ and $e_1 + e_2$. Yet its length above is $\leq 3$. How can this be?