1
here is an example: http://jsfiddle.net/LvsYc/9714/
When you select 4 different images, the 4 images are the same when showing on the screen.
HTML:
<form id="form1" runat="server">
    <input type='file' id="imgInp" multiple />
    <img id="blah" src="#" width="100" />
    <img id="blah1" src="#" width="100" />
    <img id="blah2" src="#" width="100" />
    <img id="blah3" src="#" width="100" />
</form>
JS:
 function readURL(input) {        
        if (input.files && input.files[0]) {
            var reader = new FileReader();
            reader.onload = function (e) {
                $('#blah').attr('src', e.target.result);
                $('#blah1').attr('src', e.target.result);
                $('#blah2').attr('src', e.target.result);
                $('#blah3').attr('src', e.target.result);
            }
            reader.readAsDataURL(input.files[0]);
        }
    }
    $("#imgInp").change(function(){
        readURL(this);
    });
						
There are many things you can improve there. The first is not to use different Ivs for each image. The other is you have to do a loop loop and pick up image by image. I will analyze your code and respond.
– Mauro Alexandre