La plupart des exemples sur internet utilise la méthode put pour envoyer des fichiers en upload sur un serveur web. Voici un script qui permet de le faire avec Content-Disposition: form-data.
[sourcecode language="vb"]Function UploadFile( url, file, fileType )

Set oXMLHTTP = CreateObject("MSXML2.XMLHTTP")

Dim boundary
boundary = "12345678"

Set objStream = CreateObject("ADODB.Stream")
objStream.Type = 1 ‘ binary
objStream.Open
objStream.LoadFromFile(file)

‘Stream de text
Set sOut = CreateObject("ADODB.Stream")
sOut.Charset = "us-ascii"
sOut.Type = 2 ‘ Text
sOut.Open
sOut.WriteText "–" & boundary & vbCrLf
sOut.WriteText "Content-Disposition: form-data; name=""file""; filename="""&file&"""" & vbCrLf
sOut.WriteText "Content-Type: " & fileType & vbCrLf & vbCrLf

Set sOut2 = CreateObject("ADODB.Stream")
sOut2.Charset = "us-ascii"
sOut2.Type = 2 ‘ Text
sOut2.Open
sOut2.WriteText vbCrLf & "–" & boundary & "–" & vbCrLf & vbCrLf

Set sAll = CreateObject("ADODB.Stream")
sAll.Type = 1 ‘binary
sAll.Open
sOut.Position = 0
sOut.CopyTo sAll
objStream.CopyTo sAll
sOut2.Position = 0
sOut2.CopyTo sAll

oXMLHTTP.Open "POST", url, False
oXMLHTTP.setRequestHeader "Content-Type", "multipart/form-data; boundary=" & boundary
oXMLHTTP.setRequestHeader "Connection", "close"
oXMLHTTP.setRequestHeader "Content-length", sAll.Size
sAll.Position = 0
oXMLHTTP.Send sAll.Read()
Wscript.Echo "Upload-Status: " & oXMLHTTP.statusText
End Function

UploadFile "http://upload.server/upload.php", "Configurations.zip", "application/x-zip-compressed"[/sourcecode]