Case 1: If we bet \$1 on team #1 and it wins then we will get \$2
Case 2: If we bet \$1 on team #2 and it wins then we will get \$4
Case 3: If we bet \$1 on team #3 and it wins then we will get \$6
Case 4: If we bet \$1 on team #4 and it wins then we will get \$16
Case 5: If we bet \$1 on team #5 and it wins then we will get \$21
We can bet any amount on all teams at same time. So find the correct combination (amount on each and every team) so that we have to get the profit or at least no loss regardless of which team may win.
Let $T_n$ be the fraction of our total money betted on the $n^{th}$ team.
Both Alexey Burdin and Brian Tung offered an initial solution, where $T_1=\frac{1}{2},\ T_2=\frac{1}{4},\ T_3=\frac{1}{6},\ T_4=\frac{1}{16},\ T_5=\frac{1}{21} $
See that $T_n\times W_n=1\ \ \forall n$ where $W_n$ are winnings under if $n^{th}$ team wins. That means that whatever happens our entire betting is given back. But do the fractions really sum the total? $\sum T_n = \frac{115}{112}>1$
That means that it is impossible to ensure no loss. Since you'd need to bet more than the total amount betted, what's impossible.
Knowing the probability for each team to win, though, it's possible to offer a betting that maximizes the expected return.