How's this - it goes through each paragraph (assuming each line is actually a
paragraph) and strips off the first 4 chars):

Dim x As Long
Dim strString As String

With ActiveDocument
   For x = 1 To .Paragraphs.Count
       strString = .Paragraphs(x).Range.Text
       strString = Right(strString, Len(strString) - 4)
       .Paragraphs(x).Range.Text = strString
   Next x
End With

HTH

An alternative method not using a macro would be to use a wildcard search:

Search for:  ([0-9]{3}) (*)
Replace with:  \2

(Note that there is a space after {3} and before the closing round bracket.  
{3} specifies how may digits - if there were 5 digits you'd change the 3 to a
5.  Make sure "Use wildcards" is ticked.)

By putting round brackets around any text you can "mark" it as a discrete
component that can be referred to in the replace with field as "\1" (first
bracketed section), "\2" (second bracketed section) etc.

For more information look at the (murky) help section on wildcard searches.

HTH