using foreach to iterate simultaneously through multiple lists (syntax sugar)

You can do what foreach does under the hood, but with two enumerators:

using(var e1 = list1.GetEnumerator())
using(var e2 = list2.GetEnumerator())
{
    while(e1.MoveNext() && e2.MoveNext())
    {
         var item1 = e1.Current;
         var item2 = e2.Current;

         // use item1 and item2
    }
}

For convenience, you can write an extension method like the following that takes an action:

public static void ZipDo<T1, T2>( this IEnumerable<T1> first, IEnumerable<T2> second, Action<T1, T2> action)
{
    using (var e1 = first.GetEnumerator())
    using (var e2 = second.GetEnumerator())
    {
        while (e1.MoveNext() && e2.MoveNext())
        {
            action(e1.Current, e2.Current);
        }
    }
}

and use it like:

list1.ZipDo(list2, (i1,i2) => i1.Use(i2));

By the way, you can expand this to use 3 or more lists:

public static void ZipDo<T1, T2, T3>(this IEnumerable<T1> first,
    IEnumerable<T2> second, IEnumerable<T3> third,
    Action<T1, T2, T3> action)
{
    using (var e1 = first.GetEnumerator())
    using (var e2 = second.GetEnumerator())
    using (var e3 = third.GetEnumerator())
    {
        while (e1.MoveNext() && e2.MoveNext() && e3.MoveNext())
        {
            action(e1.Current, e2.Current, e3.Current);
        }
    }
}

The approach above is required when the collections have different generic types. However, if they all have the same generic type, then you can write a flexible method that takes any number of IEnumerable<T>s:

public static void ZipAll<T>(this IEnumerable<IEnumerable<T>> all, Action<IEnumerable<T>> action)
{
    var enumerators = all.Select(e => e.GetEnumerator()).ToList();
    try
    {
        while (enumerators.All(e => e.MoveNext()))
            action(enumerators.Select(e => e.Current));
    }
    finally
    {
        foreach (var e in enumerators) 
            e.Dispose();
    }
}

and use it:

var lists = new[] {
     new[]{ 1, 1, 1 }, 
     new[]{ 2, 2, 2 }, 
     new[]{ 3, 3, 3 }};

lists.ZipAll(nums => Console.WriteLine(nums.Sum()));
// 6
// 6
// 6