Find number of triples that sum up to zero in query-intervals
Let us call the input array $A$. We build the solution in three steps:

First, for each value $v$ between $-10^6$ and $10^6$, build a list with all the indices in $i \in \{1, \dots N\}$, where $A_i = v$ and keep all these lists sorted. One way to achieve this is to iterate over $A$ and append the index of each element to its corresponding list. Let $L_v$ be the list corresponding to a value $v \in \{-10^6, \dots 10^6\}$. 

In the second step, we will build a two-dimensional array $C[N][N]$, where 
\begin{equation}
C[i][j] = \begin{cases}
|\{k: i+j+k = 0, k \in \{i, \dots j\}\}| &amp;;i\leq j,\\
0 &amp;;\text{Otherwise}.
\end{cases}
\end{equation}
We can compute $C$ by iterating over all pairs $i, j \in \{1, \dots N\}$, $j\geq i$, and for each such pair, let $u = A_i$ and $v = A_j$. Let $w = -(u + v)$. Note that $u + v + w = 0$, Using binary-search twice, find in $L_w$ the lowest index not less than $i$ and the highest index not greater than $j$. The difference of the indices of these two values plus 1 is the number of elements $w$ between $A_i$ and $A_j$, such that $A_i + A_j + w = 0$, which is exactly $C[i][j]$.

Now as a last pre-processing step, Let us compute the two-dimensional array $S$, the prefix-sums over $C$, that means:
$$
S[i][j] = C[i][j] - C[i-1][j] - C[i][j-1] + C[i-1][j-1],
$$
where we consider $C[i][j] = 0$ for $i = 0$ or $j = 0$ (since we considered 1-indexed arrays).

Now for each query $i, j$, output 
$$S[j][j] - S[i-1][j] - S[j][i-1] + S[i-1][i-1],$$
where again, we consider $S[i][j] = 0$ for $i = 0$ or $j=0$.

Based on the previous definitions, try to see why this yields the answer you are looking for.

The running time is $O(N^2\log N)$ pre-processing and $O(1)$ per query. The space complexity is $O(N^2 + M)$, where $M$ is the largest value (equal to $10^6$ in your case), which is needed to keep track of the start of each list $L_i$ for each $i \in [-M, M]$.