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The following seven pairs add to $16$ so at least one of each pair must be removed:
$(1,15)$, $(2,14)$, $(3,13)$, $(4,12)$, $(5,11)$, $(6,10)$, $(7,9)$.
 
If removing these seven is sufficient, then we would be left with $8$, $16$ and seven others.
 
But
  $16+9 = 25$      So we must remove $9$    and keep its partner $7$
    $7+2 = 9$        So we must remove $2$     and keep $14$
$14+11 = 25$      So we must remove $11$   and keep $5$
    $5+4 = 9$        So we must remove $4$     and keep $12$
$12+13 = 25$      So we must remove $13$   and keep $3$
    $3+1 = 4$        So we must remove $1$     and keep $15$
$15+10 = 25$      So we must remove $10$   and keep $6$
 
But we have kept $3$ and $6$ which add to $9$.
 
Hence it is not sufficient to remove only seven. If we remove the number $6$, we obtain a set which satisfies the condition: $\{8,16,7,14,5,12,3,15\}$ or in ascending order $\{3,5,7,8,12,14,15,16\}$.
 
Hence eight is the smallest number of numbers that may be removed.

 
 
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.