9. Among the $\mathbf{51}$ consecutive odd numbers $1,3,5, \cdots, 101$, select $k$ numbers such that their sum is $\mathbf{2013}$. The maximum value of $k$ is
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【Analysis】To maximize $k$, the selected numbers should preferably be small.
$1+3+\cdots+89=45^{2}=2025>2013$, so the selected numbers must be fewer than 45, and the sum of 44 odd numbers is even.
Therefore, the maximum value of $k$ theoretically is 43.
Now let's construct $1+3+\cdots+83=42^{2}=1764$, $2013-1764=249$, $249-101=148$, replace 83 with 99, the sum increases by 16, replace 81 with 97, the sum increases by 16, $148 \div 16=9 \cdots 4$, so 9 numbers need to be replaced, and then replace 1 more number to increase by 4.
Therefore, we can select 1 to 63, 69, and 83 to 101, which are 43 numbers.
In conclusion, the maximum value of $k$ is 43.