Here’s mine for Day 1. It’s my second implementation, after I realised during part 2 that comprehensions made the whole thing much simpler.

list =
  File.read!("input")
  |> String.trim()
  |> String.split("\n")
  |> Enum.map(&String.to_integer/1)

[{a, b} | _] = for i <- list, j <- list, i + j == 2020, do: {i, j}
IO.puts("Part1: #{a} x #{b} = #{a * b}")

[{a, b, c} | _] = for i <- list, j <- list, k <- list, i + j + k == 2020, do: {i, j, k}
IO.puts("Part2: #{a} x #{b} x #{c} = #{a * b * c}")

I’m putting my solutions on Github.

Part 1 can be optimized by spliting the list in half between greater 1010 and smaller and only searching for a pairs of one number in each. There cannot be two numbers smaller or two numbers greater than 1010 to add up to 2020.

Edit: I’ve added some benchmarks.

My solution:

defmodule AdventOfCode.Day01 do
  def part1(input) do
    {a, b} =
      input
      |> String.trim()
      |> String.split("\n")
      |> Enum.map(&String.to_integer/1)
      |> find_two_entries_that_sum_to_2020()

    a * b
  end

  def part2(input) do
    {a, b, c} =
      input
      |> String.trim()
      |> String.split("\n")
      |> Enum.map(&String.to_integer/1)
      |> find_three_entries_that_sum_to_2020()

    a * b * c
  end

  def find_two_entries_that_sum_to_2020(entries) do
    Enum.reduce_while(entries, nil, fn entry, _acc ->
      n_to_find = 2020 - entry

      case Enum.find(entries, fn x -> x == n_to_find end) do
        nil -> {:cont, nil}
        x -> {:halt, {entry, x}}
      end
    end)
  end

  def find_three_entries_that_sum_to_2020(entries) do
    try do
      for a <- entries,
          b <- entries,
          c <- entries,
          a + b + c == 2020,
          # exit as soon as we find the solution
          do: throw({:break, {a, b, c}})
    catch
      {:break, result} -> result
    end
  end
end

After finishing part 1 I realized that my approach would not work with part 2, so they’re quite different
I tried to make both parts efficient, so they stop calculation as soon as they find a correct result.