Ria writes down the numbers $$1,2,\ldots,101$$ in red and blue pens. The largest blue number is equal to the number of numbers written in blue and the smallest red number is equal to half the number of numbers written in red. How many numbers did Ria write with red pen?

Let $$b$$ be the number of blue numbers and $$r$$ be the number of red numbers. Given: $$b+r = 101$$ because every integer from $$1$$ to $$101$$ is written once.

Step 1: Identify the exact set of blue numbers
The largest blue number equals the number of blue numbers, i.e. the maximum blue integer is $$b$$. Every blue number is $$\le b$$. If any integer $$k$$ with $$1\le k \le b$$ were not written in blue, the total count of blue numbers $$\le b$$ would drop below $$b$$, and we could not compensate with a blue number $$\gt b$$ (none exist). Therefore every integer from $$1$$ to $$b$$ must be blue.
Hence the blue set is exactly $$\{1,2,\dots ,b\}$$.

Consequently, all integers $$b+1,b+2,\dots ,101$$ are red. Thus the smallest red number is $$b+1$$.

Step 2: Use the condition on the smallest red number
The smallest red number equals half the number of red numbers: $$b+1 = \frac{r}{2}\quad -(1)$$

Step 3: Relate $$r$$ and $$b$$ using the total count
From $$b+r=101\quad -(2)$$.

Step 4: Solve the two equations
From (1): $$r = 2(b+1)=2b+2$$.
Substitute in (2): $$b + (2b+2) = 101 \;\Longrightarrow\; 3b + 2 = 101 \;\Longrightarrow\; 3b = 99 \;\Longrightarrow\; b = 33.$$

Step 5: Find the required number of red numbers
$$r = 101 - b = 101 - 33 = 68.$$

Therefore, Ria wrote $$68$$ numbers with the red pen.