How to find total sum of an array then checking if its odd or even in javascript?

As the error says, you must have an initial value, 0 in our case.

function oddOrEven(array) {
  var sum = array.reduce(function(acc, intialvalue) {
    return acc + intialvalue;
  }, 0);

  if (sum % 2 == 0) {
    return "even"
  } else {
    return "odd"
  }
}

console.log(oddOrEven([1,23,4,5]))
console.log(oddOrEven([0]))
console.log(oddOrEven([1]))

No initial value means you need to set the initial value of the reduce funtion as 0.

function oddOrEven(array) {
  var sum = array.reduce(function(acc, value) {
    return acc + intialValue;
  }, 0); // <- set the reduce functions initial value

  if (sum % 2 == 0) {
    return "even";
  } else {
    return "odd";
  }
}

You can clean this code up a little by re-writing it as

const oddOrEven = (array) => {
  const sum = array.reduce((acc, value) => acc + value, 0);

  return (sum % 2 == 0) ? "even" : "odd";
};

Give an initial value to your reduce call. You could also handle the case of an empty array, here I return undefinef:

const oddOrEven = arr => arr.length
  ? arr.reduce((sum, x) => sum + x, 0) % 2 === 0 ? 'even' : 'odd'
  : undefined;

console.log(oddOrEven([1, 23, 4, 5]))
console.log(oddOrEven([0]))
console.log(oddOrEven([1]))
console.log(oddOrEven([]))

As others have pointed out, you will need to pass an initial value to reduce.

As an alternate solution, you could just count whether there are are an even or number of odd elements in the array.

const oddOrEven = (array) => array.reduce((a, i) => i % 2 ^ a, 1) ? "even" : "odd"; 

console.log(oddOrEven([1,23,4,5]))
console.log(oddOrEven([0]))
console.log(oddOrEven([1]))

Consider the following truth table:

x    y    | x+y
----------|-----
even even | even
even odd  | odd
odd  even | odd
odd  odd  | even

Bitwise & can be used to check even and odd

function oddOrEven(array) {
  var sum = array.reduce((op, inp) => op + inp, 0);
  return sum & 1 ? 'odd' : 'even' 
}

console.log(oddOrEven([1,23,4,5]))
console.log(oddOrEven([0]))
console.log(oddOrEven([1]))

For the one-line lovers

const oddOrEven = a => a.reduce((o,i) => o + i, 0) & 1 ? 'odd' : 'even' 

console.log(oddOrEven([1,23,4,5]))
console.log(oddOrEven([0]))
console.log(oddOrEven([1]))

whether the sum of array elements is even or odd and if array is empty its even.. 
function evenandOdd(arr) {
    let sumofarr = 0;
            for(i=0; i < arr.length; i++) {
        sumofarr = sumofarr + arr[i];
        }  //return sumofarr;
        if(sumofarr % 2 == 0 || arr.length == 0) 
            return 'even'; 
            else return 'odd';
} 
console.log(evenandOdd([-102, -1, 3]));