Yes, it is always possible to achieve these exact bounds. This was done in a slightly different language in

> Th. Skolem, <a href="https://www.mscand.dk/article/view/10490/8511">"On Certain Distributions of Integers in Pairs with Given Differences"</a>, Mathematica Scandinavica 5 (1957), 57–68 

For example, when $L=4k$ you can use the decomposition
<ul>
  <li>$(4k-2i,8k+i, 12k-i)$ for $0\le i\le 2k-1$</li>
  <li>$(4k-2i-1, 4k+i,8k-1-i)$ for $1\le i\le k-1$ </li>
  <li>$(2k-2i-1, 5k+2+i,7k-1-i)$ for $0\le i\le k-3$ </li>
  <li>$(1, 5k,5k+1)$ and $(4k-1, 6k+1, 10k)$</li>
</ul>

In order to translate Skolem's partition like the example I did above, write your triples in the form $(i,L+a_i, L+b_i)$ for $1\le i \le L$ and use the pairs $(a_i,b_i)$ from the paper above.