По этой блок-схеме нужно составить условие задачи

Sub z()
    Range(Cells(1, 1), Cells(100, 100)).Clear
    n1=5
    n2=3
    n3=4

    k = 1
    A = InitMatrix(n1, k, 1)
    k = k + n1 + 2
    B = InitMatrix(n2, k, 1)
    k = k + n2 + 2
    C = InitMatrix(n3, k, 1)

End Sub

Function InitMatrix(n, cx, cy)
    ReDim A(n, n)
    For i = LBound(A, 1) To UBound(A, 1)
        For j = LBound(A, 2) To UBound(A, 2)
            A(i, j) = Rnd * 200 - 100
            Cells(cx + i, cy + j) = A(i, j)
        Next
    Next

    Cells(cx, cy + n + 1) = "PositiveAverage ="
    Cells(cx, cy + n + 2) = PositiveAverage(A)

    InitMatrix = A
End Function

Function PositiveAverage(A)
    For i = LBound(A, 1) To UBound(A, 1)
        For j = LBound(A, 2) To UBound(A, 2)
            If A(i, j) > 0 Then
                s = s + A(i, j)
                k = k + 1
            End If
        Next
    Next

    PositiveAverage = s / k
End Function

Const
    n = 100;

var
    i, k_max, max, neg_max: integer;
    a: array[1..n] of integer;

begin
    for i := 1 to n do
    begin
        a[i] := random(100) - 50;
        write(a[i], ' ');
    end;
    writeln();
    
    max := a[1];
    k_max := 1;
    neg_max := 0;
    for i := 1 to n do
    begin
        if a[i] > max then begin
            max := a[i];
            k_max := 1;
        end;
        if a[i] = max then k_max := k_max + 1;
        if (a[i] < 0) and (abs(a[i]) > abs(neg_max)) then
            neg_max := a[i];
    end;
    
    writeln('Max: ', max, ' k=', k_max);
    writeln('Positive: ', max);
    writeln('Negative: ', neg_max);
end.