61. 10 distinct positive integers, where the product of any 5 is even, and the sum of all 10 numbers is odd. Then the minimum sum of these 10 positive integers is $\qquad$
Answer: 65.
Solution: Since
the product of any five numbers among the ten numbers is even,

Therefore,
there are fewer than 5 odd numbers.

Given that the sum of all numbers is odd, it can be known that there are an odd number of odd numbers among the ten numbers.
Therefore,
there are 3 or 1 odd numbers.

To make the sum of these ten numbers the smallest, if we take the positive even and odd numbers in sequence. At the same time, there should be 3 odd numbers, not 1,

because $2+4+6+8+10+12+14+1+3+5<2+4+6+8+10+12+14+16+18+1$.
In this way, the smallest sum is
$$
2+4+6+8+10+12+14+1+3+5=65
$$