Let the even numbers excluded be 2k, 2k+2, 2k+4 and 2k+ 6. Then we know that

n å i=1  i - (8k+12) = 51.5625 ×(n-4)    (1).

Hence:

[ n å i=5  i - (8k + 12)]/(n-4) ³ 51.5625 ³ [ n-4 å i=1  i]/(n-4).

It follows that:

n2 -102.125n + 392.5 ³ 0  quad (2)

and

51.5625 ³ 1 2 (n-3)    (3) .

Solving the quadratic equation n² -102.125n + 392.5 = 0 gives

n = 102.125 ±94.125 2 = 4  or 98.125

From (2) and (3) we have 98 £ n £ 106. From (1):

n2 - 102.125 n -16k + 388.5 = 0

and so

k = n2 - 102.125 n + 388.5 16 .

Using a spreadsheet to calculate k for n Î [98, 106] we find that the only integer value of k is k=11 corresponding to n=100. The unique solution to the problem is n=100 and the even numbers excluded are 22,24,26 and 28.