try the Activated Event of the form. This is fired after the Load event when
the form is activated. Ofcourse, this is also fired when the form is
activated from an inactive state. So you'll probably need to check whether
its the first time Activated has been fired (after the Load).

hope this helps..
Imran.

One way to do this is to use event delegates.
Create your procedure to show your dialog. Subscribe this to the event of
your choice (for this example I used the paint event) within the Form Load.
Within your procedure you should unsubscribe itself from the paint event so
that the dialog is only shown once.

as thus...
 private void FormLukazExample_Load(object sender, System.EventArgs e)
 {
  this.Paint +=new PaintEventHandler(OpenFileDialog);
 }

 private void OpenFileDialog(object sender, PaintEventArgs p)
 {
  this.Paint -= new PaintEventHandler(OpenFileDialog);
  System.Windows.Forms.OpenFileDialog dlg = new OpenFileDialog();
  if (dlg.ShowDialog() == DialogResult.OK)
  {
   //code
  }
 }

Hope this helps :)
Br,

Mark Broadbent.