sum_infinit.js

sum(1, 2, 3, 4) == 10 // true
sum(1, 2, 3, 4)(1, 2, 3, 4) == 20 // true
sum(1)(2)(3)(4)(1)(2)(3)(4) == 20 // true

function sum(...args) {
  function res(...args2) {
    return sum(_sum(args), _sum(args2))
  }
  // https://javascript.info/object-toprimitive
  res[Symbol.toPrimitive] = () => _sum(args)
  return res
}

function _sum(arr) {
  return arr.reduce((a,b) => a + b, 0)
}

function sum(...args) {
  const fn = sum.bind(sum, ...args);

  fn[Symbol.toPrimitive] = () => {
    return args.reduce((p, v) => p + v, 0);
  };

  return fn;
}

Golfing Nodar's solution a bit:

function sum(...args) {
  const fn = sum.bind(sum, ...args);

  fn[Symbol.toPrimitive] = () => args.reduce((p, v) => p + v);

  return fn;
}

(No need to pass 0 as the second argument to reduce when you're just adding all items -- the first item will be passed in as the first accumulator value to the reducer)

And further golfing for the heck of it:

const sum = (...args) => Object.assign(sum.bind(sum, ...args), {
  [Symbol.toPrimitive]: () => args.reduce((p, v) => p + v)
});

@atesgoral if you don't pass in 0

sum() == 1
// Uncaught TypeError: Reduce of empty array with no initial value