您需要遍历OpenFileDialog返回的文件数组。
此外，最好将代码分割成更小的单元，因为这样更容易调试和维护。
我在一个表单上放置了一个按钮和三个DataGridView控件，并使用了以下代码：

Imports System.IO

Public Class Form1

    Function GetDataFiles() As String()
        Using ofd As New OpenFileDialog() With {.InitialDirectory = "C:\Temp",
                                                .DefaultExt = ".csv",
                                                .Multiselect = True,
                                                .Title = "Select 3 files"}

            If ofd.ShowDialog = DialogResult.OK Then
                Return ofd.FileNames
            End If

        End Using

        Return Array.Empty(Of String)

    End Function

    Sub LoadData()
        Dim nFiles = 3

        Dim srcFiles = GetDataFiles().Take(nFiles)

        If srcFiles.Count <> nFiles Then
            ' Probably need to ask user to select nFiles files.
            Exit Sub
        End If

        'TODO: Maybe find a better way of getting the DGVs.
        Dim dgvs = {DataGridView1, DataGridView2, DataGridView3}

        For i = 0 To nFiles - 1
            'TODO: Possibly read the files first to get the number of columns.
            Dim nCols = 4
            Dim dt = New DataTable()
            For j = 1 To nCols
                dt.Columns.Add($"DT{i}-Col{j}")
            Next

            Dim data = File.ReadAllLines(srcFiles(i)).Where(Function(s) Not String.IsNullOrEmpty(s))

            For Each s In data
                dt.Rows.Add(s.Split(ControlChars.Tab))
            Next

            dgvs(i).DataSource = dt

        Next

    End Sub

    Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
        LoadData()

    End Sub

End Class

字符串
获取：

的数据
如果你已经将一个OpenFileDialog放到了窗体上，那么你需要选择它，按F4进入它的属性，然后在“行为”部分将“多选”属性改为“真”。