Let $a_n = n(2n+1)$.  Evaluate
\[
  \biggl | \sum_{1 \le j < k \le 36} \sin\bigl( \frac{\pi}{6}(a_k-a_j) \bigr) \biggr |.
\]
To solve the given problem, we need to evaluate the expression:
\[
\left| \sum_{1 \le j < k \le 36} \sin\left( \frac{\pi}{6}(a_k-a_j) \right) \right|
\]
where \(a_n = n(2n+1)\).

1. **Simplify the Argument of the Sine Function:**
   For any \(u \in \mathbb{Z}\), modulo \(2\pi\), we have:
   \[
   \frac{\pi}{6} \left( a_{j + 3u} - a_j \right) \equiv \frac{\pi}{2} \left( 6u^2 + 5u \right) \equiv \pi u^2 + \frac{\pi}{2} u \equiv \pi u + \frac{\pi}{2} u \equiv \frac{3\pi}{2} u \equiv -\frac{\pi}{2} u \pmod{2\pi}.
   \]
   This simplification shows that the argument of the sine function can be reduced modulo \(2\pi\).

2. **Pairing the Sine Terms:**
   We want to pair the terms \(\sin\left( \frac{\pi}{6}(a_k - a_j) \right)\) with another term such that the indices \(j\) and \(k\) are congruent modulo 3. This leads to the following notation:
   \[
   \Delta = \left| (u, v): \substack{0 \leq u < v \leq 11 \\ v - u \equiv 3 \pmod{4}} \right| - \left| (u, v): \substack{0 \leq u < v \leq 11 \\ v - u \equiv 1 \pmod{4}} \right|.
   \]
   It is given that \(\Delta = -6\).

3. **Summing the Sine Terms:**
   We now sum the sine terms over the appropriate ranges:
   \[
   \sum_{1 \le j < k \le 36} \sin\left( \frac{\pi}{6}(a_k - a_j) \right) = \sum_{1 \leq m < n \leq 3} \left( \sum_{0 \leq u < v \leq 11} \sin\left( \frac{\pi}{6}(a_{n + 3v} - a_{m + 3u}) \right) + \sum_{0 \leq v < u \leq 11} \sin\left( \frac{\pi}{6}(a_{m + 3u} - a_{n + 3v}) \right) \right).
   \]
   Using the earlier simplification, we get:
   \[
   \sum_{1 \le j < k \le 36} \sin\left( \frac{\pi}{6}(a_k - a_j) \right) = \sum_{1 \leq m < n \leq 3} \left( \sum_{0 \leq u < v \leq 11} \left( \sin\left( \frac{\pi}{6}(a_n - a_m) - \frac{\pi}{2}(v - u) \right) - \sin\left( \frac{\pi}{6}(a_n - a_m) + \frac{\pi}{2}(v - u) \right) \right) + 12 \sin\left( \frac{\pi}{6}(a_n - a_m) \right) \right).
   \]

4. **Simplifying the Sine Sum:**
   Using the identity \(\sin(x - y) - \sin(x + y) = -2 \cos(x) \sin(y)\), we get:
   \[
   \sum_{1 \le j < k \le 36} \sin\left( \frac{\pi}{6}(a_k - a_j) \right) = \sum_{1 \leq m < n \leq 3} \left( 2 \Delta \cos\left( \frac{\pi}{6}(a_n - a_m) \right) + 12 \sin\left( \frac{\pi}{6}(a_n - a_m) \right) \right).
   \]
   Given \(\Delta = -6\), we have:
   \[
   \sum_{1 \le j < k \le 36} \sin\left( \frac{\pi}{6}(a_k - a_j) \right) = -6 \left( 2 \sum_{1 \leq m < n \leq 3} \cos\left( \frac{\pi}{6}(a_n - a_m) \right) + 3 \right) + 12 \sum_{1 \leq m < n \leq 3} \sin\left( \frac{\pi}{6}(a_n - a_m) \right).
   \]

5. **Final Calculation:**
   Evaluating the sums, we find:
   \[
   \sum_{1 \le j < k \le 36} \sin\left( \frac{\pi}{6}(a_k - a_j) \right) = -18.
   \]
   Therefore, the absolute value is:
   \[
   \left| \sum_{1 \le j < k \le 36} \sin\left( \frac{\pi}{6}(a_k - a_j) \right) \right| = 18.
   \]

The final answer is \(\boxed{18}\).