Worst case 20 races, best case 13 races
Run all 25 horses in 5 races to get the following partial order, with each letter corresponding to a horse's completion time (which we assume for simplicity to be constant for each horse regardless of the number of races run).
a < b < c < d < e
f < g < h < i < j
k < l < m < n < o
p < q < r < s < t
u < v < w < x < y
Call these the initial 5 chains. The best of these are afkpu, whose relative order we don't yet know. We repeatedly race the fastest horses with unknown relative order against each other. Sometimes, fewer than 5 horses would be required, in which case we put in the next fastest of some of the chains.
Race 6 ('best of the best'): afkpu Assume without loss of generality that a < f < k < p < u . The fastest horse is then a. Remove a from the pool (of horses that must still compete) and continue.
Race 7 ('runners up'): bfcgk From the initial 5 chains and race 7, we know the next fastest horse is one of bf. The fastest horse after that would either be the other (of bf) or one of cgk; all other horses are slower than these 4. The winner and runner up of race 7 are awarded positions 2 and 3, and then removed from the pool.
Consider the initial 5 chains. In each chain, the remaining horses still form a single chain, so we still have 5 chains, though some will have fewer than 5 horses. Repeat the 'best of the best' and 'runners up' races to get 3 horses every two races (thanks @IvoBeckers). When there are at most 5 horses remaining in the pool, run them all in a single race to determine the bottom 5 positions.
Calculations: races 1-5 yield no positions, the first 21 positions are determined in the next 14 races, and the last race determines the final 4 positions. Total 5 + 14 + 1 = 20 races in the worst case.
Best case: use the above method, but assume the most convenient winning pattern possible to get more than 2 placings per race after race 6.
Race 6: afkpu, a=1.
Race 7: bfcgk, b=2,f=3,c=4. After bf are removed, the fastest 3 horses according to the established partial orders are cgk, all of which are in race 7, so we get 3 placings from this race. Assume g < k since we're looking for the best case.
Race 8: dgehk, d=5,g=6,e=7. We only really need dg for the 'best of the best', so we have room for 3 more - choose ehk. This race now also has the composition of a 'runners up' race, so we get 2 placings dg. In the best case, e also places, and all the remaining horses in the pool are slower than either h or k.
Race 9: hkilp, h=8,k=9,l=10. We don't need a 'best of the best' race since hkpu are already ordered, so this is just a 'runners up' race. In the best case, hk take out the next 2 placings, and we get a bonus placing since the remaining pool is slower than at least one of ilp.
Race 10: impjn, i=11,j=12,m=13. The 'best of the best' only needs to sort imp, so we have room for 2 more - choose jn. If ij are fastest in this race, the remaining pool is slower than at least one of mp.
Race 11: npoqu, n=14,p=15,o=16. The 'best of the best' only needs to sort np; add oqu. If np are fastest in this race, the remaining pool is slower than at least one of oqu.
Race 12: qurvs, q=17,r=18,u=19. The 'best of the best' race is not needed.
Race 13: svtwx, s=20,t=21,v=22,w=23,x=24,y=25. Last 6. Race all except one of the last in its chain (say, y). If the loser of race 13 (say x) is also in that chain, the chain places the remaining horse last (we already know x < y).
