I have an emptyListBuffer[ListBuffer[(String, Int)]]() initialized like so, and given a number n, I want to fill it with n ListBuffer[(String, Int)].

For example, if n=2 then I can initialize two ListBuffer[(String, Int)] within ListBuffer[ListBuffer[(String, Int)]]() if that makes any sense. I was trying to loop n times and use the insertAll function to insert an empty list but I didn't work.

use fill

fill is a standard Scala library function in order to fill a data structure with predefined elements. Its quite handy and save lot of typing.

ListBuffer.fill(100)(ListBuffer("Scala" -> 1))

Scala REPL

scala> import scala.collection.mutable._
import scala.collection.mutable._

scala> ListBuffer.fill(100)(ListBuffer("Scala" -> 1))
res4: scala.collection.mutable.ListBuffer[scala.collection.mutable.ListBuffer[(String, Int)]] = ListBuffer(ListBuffer((Scala,1)), ListBuffer((Scala,1)), ListBuffer((Scala,1)), ListBuffer((Scala,1)), ListBuffer((Scala,1)) ...

fill implementation in Standard library

def fill[A](n: Int)(elem: => A): CC[A] = {
    val b = newBuilder[A]
    b.sizeHint(n)
    var i = 0
    while (i < n) {
      b += elem
      i += 1
    }
    b.result()
  }

The above implementation is for one dimensional data structure.

General suggestions

Looks like you are using Scala like the Java way. This is not good. Embrace functional way for doing things for obvious benefits.

Use immutable collections like List, Vector instead of mutable collections. Do not use mutable collections until and unless you have string reason for it.

Same thing can be done using immutable List

List.fill(100)(List("scala" -> 1))

scala -> 1 is same as ("scala", 1)