Question 903078
<pre>
Let the five consecutive even integers be 

2n, 2n+2, 2n+4, 2n+6, 2n+8

Let the five consecutive odd integers be 

2k+1, 2k+3, 2k+5, 2k+7, 2k+9.

Assume, for contradiction, that their averages can be equal

If their averages are equal, then

{{{((2n)+(2n+2)+(2n+4)+(2n+6)+(2n+8))/5}}}{{{""=""}}}{{{((2k+1)+(2k+3)+(2k+5)+(2k+7)+(2k+9))/5}}}

Multiply both sides by 5

{{{(2n)+(2n+2)+(2n+4)+(2n+6)+(2n+8)}}}{{{""=""}}}{{{(2k+1)+(2k+3)+(2k+5)+(2k+7)+(2k+9)}}}

{{{2n+2n+2+2n+4+2n+6+2n+8}}}{{{""=""}}}{{{2k+1+2k+3+2k+5+2k+7+2k+9}}}

{{{10n+20}}}{{{""=""}}}{{{10k+25}}}

{{{10n-10k}}}{{{""=""}}}{{{5}}}

{{{10(n-k)}}}{{{""=""}}}{{{5}}}

{{{2(n-k)}}}{{{""=""}}}{{{1}}}

This cannot be, because the left side 
is even and the right side is odd.

This is a contradiction.

Therefore the answer is "NO", the average of 5 even integers 
can never be equal to the average of 5 odd integers.

Edwin</pre>