Use every + some:

Number.every(n => Sizes.some(obj => obj.id === n))

 Sizes = [
    {
      id: 1,
    },
    {
      id: 2,
    },
    {
      id: 3,
    },
    {
      id: 4,
    },
  ];

  Number = [1, 2, 3];
  
console.log(Number.every(n => Sizes.some(obj => obj.id === n)))

You could map the values to a new array of values to check, then check if the values includes the numbers.

const sizes = [{
    id: 1,
  },
  {
    id: 2,
  },
  {
    id: 3,
  },
  {
    id: 4,
  },
];

const check = (ns, szs) => {
  const nsToCheck = szs.map((x) => x.id); // get values
  for (const number of ns) {
    if (!nsToCheck.includes(number)) return false; // return false if !include
  }
  return true; // otherwise return true
}

const numbersWillSucceed = [1, 2, 3, 4];
const numbersWillFail = [1, 5];

console.log("Will succeed: ",check(numbersWillSucceed, sizes));
console.log("Will fail: ",check(numbersWillFail, sizes));

You can easily solve this problem with functions already existing for JavaScript and Arrays.

With:

const test1 = [
    {
      id: 1,
    },
    {
      id: 2,
    },
    {
      id: 3,
    },
    {
      id: 4,
    },
];

and:

const test2 = [1, 2, 3];

One can use:

test2.every(t2 => test1.find(t1 => t1.id === t2) !== undefined)

to check if every element from test2 exists in test1. It simply tests (every) whether every element from test1 is contained in the array test2 (with find).

Please do not use variables like "Number", since Number already exists in JavasScript, see also https://eslint.org/docs/rules/no-shadow#builtinglobals