Supposing that they are all independent Brownian bridges with $B_{j}(0)=B_{j}(T)=0$, we have for independent Brownian motions $W_{j}$

$$X(t)=\sum_{j=1}^m w_j(t)B_j(t)=\sum_{j=1}^m w_j(t)W_j(t)-\frac{t}{T}\sum_{j=1}^m w_j(t).$$

So if the weights sum to one, we get that the the first term is a Brownian motion $W(t)$ and so $X(t)=W_{t}-\frac{t}{T}$ and in turn $X_{t}$ is a Brownian bridge too.

Supposing that they are all independent Brownian bridges with $B_{j}(0)=B_{j}(T)=0$, we have for independent Brownian motions $W_{j}$

$$X(t)=\sum_{j=1}^m w_j(t)B_j(t)=\sum_{j=1}^m w_j(t)W_j(t)-\frac{t}{T}\sum_{j=1}^m w_j(t)W_j(T).$$

For the first term to be a Brownian motion $W_{t}$ we would need

$EW_{t}W_{s}=t\wedge s\sum_{j=1}^m w_j(t)w_j(s)=t\wedge s$ i.e.

$\sum_{j=1}^m w_j(t)w_j(s)=1$ for all $s,t\in [0,T]$.