encontrar el número máximo en el bucle for

You can try with following code:

public static void evenSumMax(Scanner console){
   System.out.print("How many integers?");
   int a=console.nextInt();
   int maxEven = 0;
   for(int i=1;i<=a;i++){
      System.out.print("Next integer?");
      int v=console.nextInt();
      if(v%2==0){
          if(v > maxEven)
              maxEven = v;
      }

   }
   System.out.println("Maximum even is " + maxEven);
}

You can use this: (added to your existing code)

public static void evenSumMax(Scanner console) {
    System.out.print("How many integers?");
    int a = console.nextInt();
    int sum = 0;
    int max = Integer.MIN_VALUE; // the smallest value possible in 32-bit integer
    for (int i = 1; i <= a; i++) {
        System.out.print("Next integer?");
        int v = console.nextInt();
        // assuming you're looking for max odd number in input numbers
        if (v % 2 == 0 && v > max) {
            max = v;
        }
        if (i % 2 == 0) {
            sum = sum + v;
        } else {}
    }
    System.out.println("Sum of even is " + sum);
    System.out.print("maximum even is "+ max);
}

The following will work. After checking the condition that entered value is even add it to the sum variable and compare it to the maxInt. If the entered number is larger than the previously entered number then set maxInt to new value.

 public static void evenSumMax(Scanner console){

  System.out.print("How many integers?");

  int a=console.nextInt();
  int sum=0;
  int maxInt=0;

  for(int i=0;i<a;i++)
  {

  System.out.print("Next integer?");
  int v=console.nextInt();

  if(v%2==0){
  sum=sum+v;

   if(maxInt<v){
       maxInt=v;
               }

           }
   }




  System.out.println("Sum of even is "+sum);
  System.out.println("Maximum even number is "+maxInt);


  }